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We assume that the opposite inequality holds for all $n\ge n_0$. Let $\ell=\liminf_{n\to\infty}a_n$. We have $\ell\ge 0$. We can find a subsequence $a_{k_n}\to \ell$ as $n\to\infty$, and we have $a_{k_n+1}\le a_{k_n}\sqrt[k_n+1]2-1$, but the RHS tends to $\ell-1$ as $n\to\infty$, so $a_{k_n+1}$ will eventually become $<\ell-\frac 12$, which is a contradiction. [Edit: this only works for $\ell<\infty$]

I'm not convinced ! What about ${ \ell= +\infty} $

Darn! I'll look into that .

Assume, as before, that for $n\ge n_0$ what we want to show doesn't hold anymore. We have $a_{n+k}\le S_k=a_n2^{H_{n+k}-H_n}-2^{H_{n+k}-H_{n+1}}-\ldots-2^{H_{n+k}-H_{n+k-1}}-1$, where $H_n=1+\frac 12+\ldots+\frac 1n$. Let's fix $n$ and make $k\to\infty$. It would be enough to show that $S_k\to-\infty$ to obtain a contradiction. Clearly, $2^{H_{n+k}-H_n}\to\infty$ as $k\to\infty$, so it's enough to show that $\frac{2^{H_{n+k}-H_{n+1}}+\ldots+2^{H_{n+k}-H_{n+k-1}}+1}{2^{H_{n+k}-H_n}}\to\infty$ as $k\to\infty$ (remember, $n$ is fixed). We can multiply that fraction by $2^{H_n}$, which is constant, to get a nicer expression. We can also add the constant term $2^{-H_1}+\ldots+2^{-H_n}$, to get, in the end, $\sum_{t\ge 1} 2^{-H_t}=\infty\ (*)$ (this is what we want to prove). Since $\sum_{t\ge 1}2^{-\ln t}=\sum_{t\ge 1}\frac 1{t^{\ln 2}}=\infty$, we can use Cesaro-Stolz to prove $(*)$. It looks rather nasty. Does it work?

Basic calculus shows that $ 2^{\frac{1}{n}}\le 1+\frac{1}{n}=\frac{n+1}{n}$ for all $ n\ge 1$. Thus, we need to show there exists no sequence such that for all $ n\ge k$, $ 1+a_n\le a_{n-1}(2^\frac{1}{n})\le a_{n-1}\frac{n+1}{n}$. Recursively, we can find that $ a_{n-1+m}\le a_{n-1}\frac{n+m}{n}-\frac{n+m}{n+1}-\frac{n+m}{n+2}-\cdots -\frac{n+m}{n+m}$. It suffices, to get a contradiction, to show that the RHS is eventually negative, i.e. $ a_{n-1}\frac{n+m}{n}\le \frac{n+m}{n+1}+\cdots +\frac{n+m}{n+m}$ for some large finite $ m$. Dividing out by $ n+m$ we get that $ \frac{a_{n-1}}{n}\le \frac{1}{n+1}+\cdots \frac{1}{n+m}$, which is true for some finite $ m$ by the divergence of the harmonic series.

GoldenFrog1618 wrote: Basic calculus shows that $ 2^{\frac{1}{n}}\le 1+\frac{1}{n}=\frac{n+1}{n}$ for all $ n\ge 1$. Alternatively, $\left(1 + \frac{1}{n}\right)^n = \sum_{k=0}^n \frac{\binom{n}{k}}{n^k} = 2 + \sum_{k=2}^n \frac{\binom{n}{k}}{n^k} \geq 2$ (where $n < 2$ implies an empty sum.)

On the contrary we assume its true for finitely many $n$, hence after some finite terms (say $N$ ) it no longer holds. We have $a_{N}\leq 2^{\frac{1}{N}}a_{N-1}-1\leq 2^{\frac{1}{2}}a_{N-1}-1$. Hence we may safely assume a new sequence $b_i := a_{i+N}$. We have $b_n\leq 2^{\frac{1}{n}}b_{n-1}-1$. Let $S_n=2^{1+\cdots+\frac{1}{n-1}+\frac{1}{n}}+2^{\frac12+\cdots+\frac{1}{n-1}+\frac{1}{n}}+\cdots+2^{\frac{1}{n}}+1$. Note that we have \[b_n\leq 2^{H_n}b_0-S_n\] I shall be proving that \[\frac{S_n}{2^{H_n}}>\frac{S_{n-1}}{2^{H_{n-1}}}\]\[\iff S_n>2^{\frac{1}{n}}S_{n-1}\] which is obviously true. Thus $b_0-\frac{S_n}{2^{H_n}}<0$ after some time and we get a contradiction.

I solved this problem a long time before,probably this was my solution: First of all,by Bernoulli inequality we have $(1+\frac{1}{n})^n \ge 2 \Rightarrow 1+\frac{1}{n} \ge 2^{\frac{1}{n}}$ Thus it is enough to show that $1+a_n > (1+\frac{1}{n})a_{n-1}$ for infinitely many $n$. We assume just the opposite for the sake of contradiction,i.e,there exists an $M$ such that $\forall n \ge M$ we have $1+a_n \le (1+\frac{1}{n})a_{n-1}$ or $\frac{1}{n+1}+\frac{a_n}{n+1} \le \frac{a_{n-1}}{n}$.Summing this for $M,M+1,\cdots,m$ we obtain $\frac{a_{m-1}}{m} \le \frac{a_{M-1}}{M}-(\frac{1}{m}+\frac{1}{m+1}+\cdots+\frac{1}{M+1})$.But it is a well known consequence of AM-HM inequality that the harmonic sum on the RHS increases without bound,so letting $m \rightarrow \infty$ we get $a_{m-1}<0$.This is a contradiction since all the numbers in the sequence are positive.

As grobber pointed out, the problem is solved if we can show that $\sum_{k=1}^{\infty}(\frac{1}{2})^{H_k}$ diverges, where $H_k$ is the $k$th harmonic number. This can be done with some stupid bounds. We first give a upper bound for $H_k$. If $2^a \le i <2^{a+1}$ then $\frac{1}{i}\le \frac{1}{2^a}$ so $\sum_{i=2^a}^{2^{a+1}-1}\frac{1}{i} \le 1$. It follows that $H_{2^k-1} \le k$ for $k \ge 1$. Then if $2^{k-1} \le i < 2^k$ then $H_i \le H_{2^k-1} \le k$ so $(\frac{1}{2})^{H_i} \ge (\frac{1}{2})^{H_{2^k-1}} \ge (\frac{1}{2})^k$. So the sum is at least $\sum_{k=1}^{\infty}2^{k-1}(\frac{1}{2})^k=\sum_{k=1}^{\infty}\frac{1}{2}$ which obviously diverges.

The problem seems a bit easy so probably my solution is wrong ... Do check and point out Solution : Let there be finitely many such $n$. Let $m-1$ be the largest integer such that $1+a_{m-1} > a_{m-2} \sqrt[m-1]{2}$ \[\implies 1+a_n \le a_{n-1}\sqrt[n]{2} \] $\forall n \ge m$ \[1+a_m \le a_{m-1}\sqrt[m]{2} \] \[1+a_{m+1} \le a_{m}\sqrt[m+1]{2} \] \[ \cdots\] \[1+a_{m+k} \le a_{m+k-1}\sqrt[m+k]{2} \] Adding all of them, \[k+\sum _{i=m}^{m+k}a_i \le \sum _{i=m}^{m+k}a_{i-1}\sqrt[i]{2}\] But this is clearly not possible for a sufficiently large choice of $k$ and since $\sqrt[j]{2}<\sqrt{2}$ (use AM-GM) Thus a contradiction. Hence there must exist infinitely many such $n$

utkarshgupta wrote: But this is clearly not possible since $\sqrt[j]{2}<1$ and $a_{m-1}<k$ Nope. First, there doesn't have to exist such $m$ (if the inequality holds for all $n$), but more importantly: $\sqrt[j]{2}>1$!

I assumed there are finitely many such $n$ so there must exist a largest ... And as for the second mistake, i have corrected it so please relook Thanks

Note that by Bernoulli's inequality, \[ (1+\frac{1}{n})^{n}\ge 2\Rightarrow \sqrt[n]{2}\le 1+\frac{1}{n} \]for each $n\ge 1$. Now suppose that there is some $m$ with $a_{n}\le a_{n-1}\sqrt[n]{2}-1$ for all $n>m$. By our previous estimate, this implies \[ a_{n}\le a_{n-1}\frac{n+1}{n}-1 \]for each $n>m$. So \begin{align*} a_{m+1}&\le a_{m}\frac{m+2}{m+1}-1\\ \Rightarrow a_{m+2}&\le a_{m+1}\frac{m+3}{m+2}-1\le a_m\frac{m+3}{m+1}-\frac{m+3}{m+2}-1\\ \Rightarrow a_{m+3}&\le a_{m+2}\frac{m+4}{m+3}-1\le a_m\frac{m+4}{m+1}-\frac{m+4}{m+2}-\frac{m+4}{m+3}-1 \end{align*}and in general, for each $k\ge 1$, \[ a_{m+k}\le (m+k+1)\left(\frac{a_m}{m+1}-\frac{1}{m+2}-\frac{1}{m+3}-\cdots - \frac{1}{m+k+1}\right). \]But as $k\to \infty$, the second term tends to $O(1)-O(\log k)$, which is a contradiction if $a_k>0$ always.

Suppose that there existed an integer $N$ such that for all $n>N$ the inequality was false. Let $b_n=a_n+\sum_{i=1}^{n}2^{H_n-H_i}$, where $H_k=\sum_{i=1}^{k}\frac{1}{i}$. Note that the condition now becomes $b_n=2^\frac{1}{n}b_{n-1}$, so we have that $b_m=2^{H_m-H_N}b_N=c2^{H_m}$ for some constant $c$, so $a_m=c2^{H_m}-\sum_{i=1}^{m}2^{H_m-H_i}$. We claim that $\lim_{m\rightarrow\infty}\frac{\sum_{i=1}^{m}2^{H_m-H_i}}{2^{H_m}}=\infty$, which will imply that for some $m$ $\frac{\sum_{i=1}^{m}2^{H_m-H_i}}{2^{H_m}}>c$ so $a_m<0$. But this is equivalent to $\sum2^{-H_i}$ diverging. Since $H_i<\log i$ by integration, $\sum2^{-H_i}>\sum2^{-\log i}=\sum e^{-\log 2\log i}=\sum i^{-\log 2}>\sum i^{-1}$ which diverges as desired.

Routine Approximations: Suppose for sake of contradiction that for all $m\ge n$, we have $1+a_m\le a_{m-1}\sqrt[m]{2}$. Lemma 1: We have $a_{n+N}\le a_n2^{\frac{1}{n+1}+\cdots+\frac{1}{n+N}}-2^{\frac{1}{n+2}+\cdots+\frac{1}{n+N}}-\cdots-2^{\frac{1}{n+N}}-1$. Proof of Lemma 1: Iterate the inequality $a_m>a_{m-1}2^{1/m}-1$ $N$ times. For small values of $N$, we have \begin{align*} a_{n+1}&\le a_n 2^{\frac{1}{n+1}}-1 \\ a_{n+2}&\le\left(a_n 2^{\frac{1}{n+1}}-1\right)2^{\frac{1}{n+2}}-1 \\ a_{n+3}&\le\left[\left(a_n 2^{\frac{1}{n+1}}-1\right)2^{\frac{1}{n+2}}-1\right]2^{\frac{1}{n+3}}-1 \\ &\,\,\vdots \end{align*} Proof of Lemma 2: $\log(b/a)-1<\frac{1}{a+1}+\cdots+\frac{1}{b}<\log(b/a)+1$ for positive integers $a,b$. Proof of Lemma 2: Follows from bounding $\int_a^b\frac{1}{x}\,dx$ by left and right Riemann sums. By applying lemma 1 and lemma 2 many times, we see that \begin{align*} a_{n+N} &\le a_n 2^{\log\left(\frac{n+N}{n}\right)+1}-\left[2^{\log\left(\frac{n+N}{n+1}\right)-1}+\cdots+2^{\log\left(\frac{n+N}{n+N}\right)-1}\right] \\ &= (n+N)^{\log 2}\left[\frac{2a_n}{n^{\log 2}}-\frac{1}{2}\left[(n+1)^{-\log 2}+\cdots+(n+N)^{-\log 2}\right]\right]. \end{align*}But we know that $a_{n+N}>0$, so we must have that \[(n+1)^{-\log 2}+\cdots+(n+N)^{-\log 2}<\frac{4a_n}{n^{\log 2}}\]for all $N$. But Since $\log 2<1$, we then have that the left side diverges to $+\infty$ as $N\to\infty$ by series comparison with $(n+1)^{-1}+(n+2)^{-1}+\cdots$. Therefore, it cannot be bounded above, so we have the desired contradiction.

Really good problem. Assume by way of contradiction that $N$ was the largest value for which this holds. Then for all $k> N$, we have $1+a_k\le a_{k-1}\sqrt[k]{2}$. First, we may show by induction that $$1+a_{N+m}\le a_N2^{1/(N+1)+\cdots +1/(N+m)}-\left(2^{1/(N+2)+\cdots+1/(N+m)}+2^{1/(N+3)+\cdots+1/(N+m)}+\cdots +2^{1/(N+m)}\right)$$Therefore, the RHS is greater than $1$. This implies \begin{align*} a_N &>\frac{ 1+2^{1/(N+2)+\cdots+1/(N+m)}+2^{1/(N+3)+\cdots+1/(N+m)}+\cdots +2^{1/(N+m)} }{2^{1/(N+1)+\cdots +1/(N+m)}}\\ &=\frac{1}{2^{1/(N+1)}}+\frac{1}{2^{1/(N+1)+1/(N+2)}}+\cdots+\frac{ 1 }{2^{1/(N+1)+\cdots +1/(N+m)}} \end{align*}Standard calculus techniques show that $$\frac{1}{a}+\frac{1}{a+1}+\cdots +\frac{1}{b}\le 1+\ln\left(\frac{b}{a}\right)\le 1+\log_2\left(\frac{b}{a}\right).$$Therefore, \begin{align*} a_N &>\frac{1}{2^{1/(N+1)}}+\frac{1}{2^{1/(N+1)+1/(N+2)}}+\cdots+\frac{ 1 }{2^{1/(N+1)+\cdots +1/(N+m)}}\\ &\ge \frac{N+1}{2(N+1)}+\frac{N+1}{2(N+2)}+\cdots+\frac{N+1}{2(N+m)}\\ &= \frac{N+1}{2}\left(\frac{1}{N+1}+\frac{1}{N+2}+\cdots+\frac{1}{N+m}\right)\end{align*}But as $m$ approaches infinity, the RHS diverges, contradiction since $a_N<\infty$.

For contradiction, $\exists N>0$ such that $ 1+a_n>a_{n-1} \sqrt[n]{2} \forall n \ge N$ We have $$ 1+a_n \le 2^{\frac{1}{n}}a_{n-1} \forall n le N$$Let $(x_n)$ such that $$\left\{ \begin{matrix} {{x}_{1}}=1 \\ {{x}_{n}}={{2}^{\frac{-1}{2}-\frac{-1}{3}-.....-\frac{1}{n}}},\forall \,n\ge 2 \\ \end{matrix} \right.$$Then, we have $$ x_n+x_na_n \le x_{n-1}a_{n-1}, \forall n \ge N$$So $$ x_{N-1}a_{N-1} \ge x_{N+t}a_{N+t}+x_N+x_{N+1}+\ldots+x_{N+t} (1). $$We have $x_N+x_{N+1}+\ldots+x_{N+t} \to +\infty$ when $t\to\infty$, contradiction

Suppose for the sake of contradiction there exists some $N$ such that for all $n\ge N$, we have $1+a_n \le 2^{1/n}\cdot a_{n-1}$. It is easy to show by induction that for all $k\ge 0$, we have \begin{align*} a_{N+k} &\le \left( 2^{ \tfrac{1}{N} + \tfrac{1}{N+1} + \cdots + \tfrac{1}{N+k}}\right)a_{N-1} - \left( 1 + \sum_{i=1}^k 2^{ \tfrac{1}{N+i} + \cdots + \tfrac{1}{N+k}} \right) \end{align*}Let $C=a_{N-1}$ be a constant. We want to find some $k$ for which $a_{N+k}$ is negative, as this produces a contradiction. By the above, $a_{N+k}<0$ if \[ 1 + \sum_{i=1}^k 2^{ \tfrac{1}{N+i} + \cdots + \tfrac{1}{N+k}} > C\cdot 2^{ \tfrac{1}{N} + \tfrac{1}{N+1} + \cdots + \tfrac{1}{N+k}}. \qquad (\heartsuit) \]We will find a $k\le N$ satisfying the above. For $0<\varepsilon<1$, one can show $2^\varepsilon >1+0.01\varepsilon$. Since $\tfrac{1}{N}+\cdots+\tfrac{1}{N+k}<1$ for $k\le N$, we can lower bound the LHS of $(\heartsuit)$: \begin{align*} 1 + \sum_{i=1}^k 2^{ \tfrac{1}{N+i} + \cdots + \tfrac{1}{N+k}} &> (k+1) + 0.01\sum_{i=1}^k \left( \frac{1}{N+i} + \cdots + \frac{1}{N+k} \right) \\ &= (k+1) + 0.01\left[ \frac{1}{N+1}+\frac{2}{N+2}+\cdots+\frac{k}{N+k} \right] \\ &= (1.01k+1) - 0.01N\left[\frac{1}{N+1} + \frac{1}{N+2}+\cdots + \frac{1}{N+k} \right]. \end{align*}Similarly, for $0<\varepsilon<1$, one can show that $2^\varepsilon < 1+\varepsilon$, so we can upper bound the RHS of $(\heartsuit)$: \[ C\cdot 2^{ \tfrac{1}{N} + \tfrac{1}{N+1} + \cdots + \tfrac{1}{N+k}} < C \left( 1 + \frac{1}{N} + \frac{1}{N+1} +\cdots+\frac{1}{N+k}\right). \]Therefore, it suffices to find a $k\le N$ for which \[ (1.01k+1) - 0.01N\left[\frac{1}{N+1} +\cdots + \frac{1}{N+k} \right] > C \left( 1 + \frac{1}{N} +\cdots+\frac{1}{N+k}\right),\]i.e. \begin{align*} 1.01k+1-C-\frac{C}{N} &> (C+0.01N)\left( \frac{1}{N+1} +\cdots + \frac{1}{N+k} \right) \\ &= (C+0.01N)(H_{N+k}-H_N). \end{align*}Take $k=N$. Then $H_{N+k}-H_N = H_{2N}-H_N < \ln(2)$, and since $1.01N-C-C/N > (C+0.01N)\ln(2)$ for large enough $N$, we are done.

Fakesolve or not? I can't find an error, but the other solutions seem longer and I can't find a similar solution to mine. Suppose that the conclusion is not true. Then there exists a positive integer $N$ such that for all $n>N$, $1 + a_n \le a_{n-1} \sqrt[n]{2}$. Define a new sequence $b_n = \frac{a_n}{n}$, and notice that this is always positive. By Bernoulli's inequality $\left(1+\frac{1}{n}\right)^n>2$ and thus $1+\frac{1}{n} > \sqrt[n]{2}$. Then notice that for all $n>N$, \begin{align*} b_n &= \frac{a_n}{n} \\ & \le \frac{a_{n-1}\sqrt[n]{2}-1}{n} \\ & = \frac{(n-1)b_{n-1}\sqrt[n]{2}-1}{n} \\ & <\frac{(n-1)b_{n-1}\left(1+\frac{1}{n}\right)-1}{n} \\ &= \frac{\frac{n^2-1}{n}b_{n-1}-1}{n} \\ &= b_{n-1} - \frac{b_{n-1}}{n^2}-\frac{1}{n} \\ &\le b_{n-1}-\frac{1}{n}.\end{align*}This implies that for $x>N$, $$b_x < b_N - \sum_{i=N+1}^x \frac{1}{i}.$$As the harmonic series diverges, for sufficiently large $x$, $b_x$ will be negative, contradiction.

Just like #21, this seems a lot shorter (in idea, not actual post length) compared to the others on the thread. The problem is almost two separate problems put together but the conclusion is relatively cool nevertheless.

Assume that there exists a maximum $n$ for which the inequality holds. For all $n\ge k, 1 + a_n \le a_{n-1}\sqrt[n]{2}$ Note that by Bernoulli's inequality, $\sqrt[n]{2}\le 1+\frac{1}{n},$ thus, \[1+a_n\le \frac{n+1}{n}a_{n-1}\]for all $n\ge k.$ Then, \[a_{k+i}\le \frac{k+i+1}{k}a_{k-1}-\frac{k+i+1}{k+1}-\frac{k+i+1}{k+2}-\frac{k+i+1}{k+3}-\ldots-\frac{k+i+1}{k+i+1}\]We claim that for sufficiently large $i$ this number is negative. Note that this is equivalent to proving that $\frac{1}{k+1}+\frac{1}{k+2}+\dots+\frac{1}{k+i+1}$ diverges, which is true.

I think you may have posted this in the wrong thread.

Assume for the sake of contradiction that the inequality is only statisfied for finitely many $n$. That is, there exists an index $N$ such that $$a_n \leq a_{n-1}\sqrt[n]{2}-1$$for all $n\geq N$. Note that $\sqrt[n]{2}\leq 1+\frac{1}{n}$, so we have $a_n \leq a_{n-1}(1+\frac{1}{n})-1$ for all $n\geq N$. We then have $a_{N+1}\leq 2a_N-1$, $a_{N+2}\leq 3a_N-\frac{5}{2}$, $a_{N+3}\leq 4a_N-\frac{26}{6}$, and so on. In general, we have $a_{N+k}\leq (k+1)a_N-\frac{S_k}{k!},$ where $S_k$ denotes the $k$th Sterling number. This can be shown by induction, as that term multiplies by $\frac{k+1}{k}$ and subtracts 1 each time. It is well known that $\frac{S_k}{k!}$ eventually dominates any linear function, so this eventually becomes negative, contradiction.

Assume not. Then there is a number $k$ such that $$a_n \le a_{n-1} \sqrt[n]{2} -1,$$for all $n \ge k$. Using the binomial theorem, we get $\sqrt[n]{2} \le 1 + \frac{1}{n}$, so $a_n \le a_{n-1} \left( 1 + \frac{1}{n} \right) -1$ for all $n \ge k$. Let $b_n := \frac{a_n}{n+1}$. Using this notation we get $$b_n \le b_{n-1}-\frac{1}{n+1} \implies 0 \le b_n \le b_k -\frac{1}{k+2}-\frac{1}{k+3} - \cdots - \frac{1}{n+1} \implies \frac{1}{k+2}+\frac{1}{k+3} + \cdots +\frac{1}{n+1} \le b_k.$$We know that the harmonic series is unbounded and so taking big enough $n$ produces a contradiction and we are done.

Suppose otherwise. Then there exists some constant $c$ such that $1+a_n\le a_{n-1}\sqrt[n]{2}$ for all $n\ge c$. Fix $a_c$ and consider the sequence $b_c,b_{c+1},\dots$ such that $b_c=a_c$ and $1+b_n=b_{n-1}\sqrt[n]{2}$. I claim that $b_n\ge a_n$. We show this by induction. Clearly the base case $n=c$ holds. Then, suppose that $b_n\ge a_n$. Then $a_{n+1}\le a_n\sqrt[n+1]{2}-1\le b_n\sqrt[n+1]{2}-1=b_{n+1}$. Now note that it suffices to prove the case where $c=0$; for all other cases, we can simply "work backwards" until we reach the $c=0$ case by constructing the terms $b_{c-1},b_{c-2},\dots$. Associate with each term $b_n$ a pair of numbers $c_{n,1}$ and $c_{n,2}$ such that $b_n=c_{n,1}b_0-c_{n,2}$. Then note that it suffices to show that $\frac{c_{n,2}}{c_{n,1}}$ is unbounded to achieve a contradiction. Note that \[c_{n,1}=2^{\tfrac{1}{1}+\tfrac{1}{2}+\dots+\tfrac{1}{n}}\]\[c_{n,2}=2^{\tfrac{1}{2}+\dots+\tfrac{1}{n}}+2^{\tfrac{1}{3}+\dots+\tfrac{1}{n}}+\dots+2^{\tfrac{1}{n}}+2^0\]so it is enough to show that \[\frac{1}{2^{\tfrac{1}{1}}}+\frac{1}{2^{\tfrac{1}{1}+\tfrac{1}{2}}}+\dots+\frac{1}{2^{\tfrac{1}{1}+\tfrac{1}{2}+\dots+\tfrac{1}{n}}}\]is unbounded. Then it suffices to show that \[\sqrt[n]{2}\le 1+\frac{1}{n}\]since this will imply by induction that \[2^{\tfrac{1}{1}+\tfrac{1}{2}+\dots+\tfrac{1}{n}}\le n+1.\]But now by Bernoulli's Inequality we are done. $\blacksquare$

Suppose the problem is not true, then there exists $N\in \mathbb{N}$ such that for all $n>N$ we have $1+a_n\leq a_{n-1}\sqrt[n]{2}$ By an induction we can show that $a_{N+k}\leq a_N2^{\frac{1}{N+k}+...+\frac{1}{N+1}}-\sum_{i=0}^{k}2^{\frac{1}{N+k}+...+\frac{1}{N+k+1-i}}$ Let $H_n=\sum_{i=1}^{n}\frac{1}{i}$ be the $n$-th harmonic number ($H_0=0$) Then $a_{N+k}\leq a_N2^{H_{N+k}-H_N}-\sum_{i=0}^{k}2^{H_{N+k}-H_{N+i}}=f(k)$ Using inequality $H_n<\ln n+1$ we get $f(k)=2^{H_{N+k}}(\frac{a_N}{2^{H_N}}-\sum_{i=0}^{k}\frac{1}{2^{H_{N+i}}})<2^{H_{N+k}}(\frac{a_N}{2^{H_N}}-\sum_{i=0}^{k}\frac{1}{2^{\ln(N+i)+1}})=2^{H_{N+k}}(\frac{a_N}{2^{H_N}}-\frac{e}{2}\sum_{i=0}^{k}\frac{1}{N+i})=2^{H_{N+k}}(\frac{a_N}{2^{H_N}}-\frac{e}{2}(H_{N+k}-H_{N-1}))$ We want $f(k)<0$ so we want $\frac{a_N}{2^{H_N}}-\frac{e}{2}(H_{N+k}-H_{N-1})<0$ This is equivalent to $H_{N+k}>\frac{2a_N}{2^{H_N}e}+H_{N-1}$ but this holds true for sufficiently large $k$ Hence for sufficiently large $k$ we have $a_{N+k}<0$, contradiction

It's obvious that $(1+\frac{1}{n})^{n}\ge 2\implies\sqrt[n]{2}\le 1+\frac{1}{n}$. Then, take the largest number $a_l$ s.t. the inequality still holds (FTSOC). It follows that $$a_{l+1}\le\sqrt[l]{2}a_l-1\le a_l\frac{l+2}{l+1}-1\implies a_{l+2}\le a_{l+1}\frac{l+3}{l+2}-1\le a_l\frac{l+3}{l+1}-\frac{l+3}{l+2}-1,$$which, by a simple induction, implies $$a_{l+k}\le(l+k+1)\left(\frac{a_l}{l+1}-\frac{1}{l+2}-\frac{1}{l+3}-\dots-\frac{1}{l+k+1}\right);$$in particular, for sufficiently large k note that 1/1+1/2+...+1/n is unbounded, so even without the 1/1+1/2+...+1/{l+1} this sequence tends to infinity, which implies $\left(\frac{a_l}{l+1}-\frac{1}{l+2}-\frac{1}{l+3}-\dots-\frac{1}{l+k+1}\right)$ becomes negative, but this is a contradiction because all terms in the sequence are positive. $\blacksquare$ this should work but idk those types of wonky stuff im notrelaly good at also the induction is provided for detail: Base case is done, inductive step: $$a_{l+k+1}\le \frac{l+k+2}{l+k+1}a_{l+k}-1\le(l+k+2)\left(\frac{a_l}{l+1}-\frac{1}{l+2}-\frac{1}{l+3}-\dots-\frac{1}{l+k+1}-\frac1{l+k+2}\right)$$

we are now engaging in the toxic completionist desire of writing up every problem that is done. this is making myself pay for 540 500. Claim: It suffices to show that there is at least one index $i$ satisfying $a_{i + 1} + 1 > \sqrt[i+1]{2} a_{i} $. Proof: Basically once we get that one element, consider the sequence starting from $a_{i + 1}$, any future elements $j$ satisfying $a_j + 1 > \sqrt[j - i]{2}a_{j-1}$ (basically this is the condition starting the indexing at $i + 1$) also satisfy the condition when the root is replaced with $j$ (basically this is the condition starting the indexing at $1$). Claim: There is at least one index $i$ satisfying the condition. Proof: Assume not, then we have $a_2 \le 2a_1 - 1, a_3 \le \sqrt{2}a_2 - 1, \cdots$. This then implies $0 < a_n \le 2^{1 + \frac 12 + \frac 13 + \cdots} a_1 - 2^{\frac 12 + \frac 13 + \cdots \frac 1n} - 2^{\frac 13 + \frac 14 + \cdots \frac 1n} - \cdots - 2^{\frac 1n} - 1$. We claim that this is impossible, in fact that the expression on the right is eventually negative. This is equivalent to proving that $ 2^{\frac 12 + \frac 13 + \cdots \frac 1n} + 2^{\frac 13 + \frac 14 + \cdots \frac 1n} + \cdots + 2^{\frac 1n} + 1 \ge 2^{1 + \frac 12 + \frac 13 + \cdots} a_1 $. We can instead prove the much weaker statement that $n - 1 \ge 2^{H_n}$ for large enough $n$, this is because the left side is $O(n)$ and the right side is $O(2^{\ln x})$ (trivial by integral) which is $O(x^{\ln 2})$ so we are done.

Note that it suffices to show that if $a_n=m$, then it holds for some $k>n$ for all $n,m$. Therefore, suppose not, so that $a_n\le a_{n-1}\sqrt[n]{2}-1$. Our goal is to prove that $a$ eventually goes to or below $0$, so we can make the $\le$ an equality. Note that multiplying by $\sqrt[n]{2}$ then subtracting $1$ will decrease the number if it is below $b_n=\frac{1}{\sqrt[n]{2}-1}$. Claim. Eventually, $b_n>a_n$. Proof. Note that \[b_n=\frac{1}{\sqrt[n]{2}-1}=\frac{1/2}{\sqrt[2n]{2}-1}-\frac{1/2}{\sqrt[2n]{2}+1}\sim\frac12 b_{2n}-\frac14.\]Therefore, the $b_{2n}\sim 2b_n$. However, \[a_n\le a_{n-1}\sqrt[n]{2},\]so \[a_{2n}\le 2^{\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{2n}}\cdot a_n\sim 2^{\ln 2}a_n\ll 2a_n.\]Therefore, $b$ grows much faster than $a$. $\blacksquare$ Now note that when $a_k<b_k$ and $k>n$, \[a_k=a_{k-1}\sqrt[k]{2}-1\le a_{k-1}\sqrt[n]{2}-1\Longrightarrow b_n-a_k\ge (b_n-a_{k-1})\sqrt[n]{2}.\]Thus the gap between $b_n$ and $a_k$ widens a diverging amount so we win. $\blacksquare$

We claim for any nonnegative integer $n$, there always exists an integer value of $k > n$ where the inequality holds. Assume otherwise. Observe that $a_{n+1} \leq a_n \sqrt[n+1]{2} - 1$, $a_{n+2} \leq (a_n \sqrt[n+1]{2} - 1) \sqrt[n+2]{2} - 1$, and so on. Expanding gives the general form $$a_m \leq 2^{\frac{1}{n+1} + \frac{1}{n+2} + \dots + \frac{1}{m}} a_n - 2^{\frac{1}{n+2} + \frac{1}{n+3} + \dots + \frac{1}{m}} - 2^{\frac{1}{n+3} + \frac{1}{n+4} + \dots + \frac{1}{m}} - \dots - 2^{\frac{1}{m}} - 1$$which is equivalent to $$2^{-\frac{1}{n+1} - \frac{1}{n+2} - \dots - \frac{1}{m}} a_m \leq a_n - 2^{-\frac{1}{n+1}} - 2^{-\frac{1}{n+1} - \frac{1}{n+2}} - \dots - 2^{-\frac{1}{n+1} - \frac{1}{n+2} - \dots - \frac{1}{m}}$$Let $S_i = \sum_{j=n+1}^i \frac{1}{j}$ for all $i > n$; we claim that $\sum_{i=n+1}^m 2^{-S_i}$ diverges. WLOG assume $n = 0$; then for every positive integer $p$, $S_i \leq p$ for all $2^{p-1} \leq i < 2^p$, and $\sum_{i=2^{p-1}}^{2^p-1} 2^{-S_i} \geq \frac{1}{2}$. As a result, taking $m = n + 2^{2a_n}$ guarantees that $\sum_{i=n+1}^m 2^{-S_i} \leq a_n$, leading to $a_m \leq 0$ and therefore a contradiction.

Otherwise, $1+a_n<2^{1/n}a_{n-1}$ for all sufficiently large $n>m$; WLOG assert equality. Intuitively, $\{a_i\}$ should eventually dip below $0$. Rigorously, \begin{align*} a_n&=2^{1/n}a_{n-1}-1\\ &=2^{1/n}(2^{1/(n-1)}a_{n-2}-1)-1\\ &\vdots\\ &=2^{1/n}(2^{1/(n-1)}(\dots(2^{1/(m+1)}a_m-1)\dots)-1)-1, \end{align*}which expands as \[2^{1/(m+1)+\dots+1/n}a_m-1-2^{1/(m+1)}-\dots-2^{1/(m+1)+\dots+1/n}.\]Approximating $\textstyle\sum_{a+1}^b\tfrac{1}{x}=\int_a^b\tfrac{1}{x}=\ln\tfrac{b}{a}$ gives \[a_n\approx2^{\ln n/m}a_m-2^{\ln m/m}-\dots-2^{\ln n/m}.\]We seek sufficiently large $n$ such that the above is negative, i.e., \[2^{\ln n/m}a_m<\sum_{i=m}^n2^{\ln i/m}\iff a_m<2^{1/n}\sum_{i=m}^n2^{\ln i}.\]Again, we approximate $\textstyle\sum2^{\ln i}\approx\int2^{\ln i}=\tfrac{x^{1+\ln2}}{1+\ln2}$ to find \[a_m<2^{1/n}(\tfrac{n^{1+\ln2}}{1+\ln2}-\tfrac{m^{1+\ln2}}{1+\ln2}),\]which diverges for $m\mapsto\infty$ since $1+\ln2>0$ (!).

Lemma: Let $H_n$ be the nth harmonic number. Then $n>H_n$ whenever $n>4$. Proof. We can check this for $n=5$. Now suppose $m>2^{H_m}$. Notice that $$m+1-2^{H_{m+1}}=m+1-2^{H_m} \cdot 2^{1/(m+1)}>m+1-m2^{1/(m+1)}>0$$where the last inequality is true because $(m+1)^{m+1}>2\cdot m^{m+1}$. Now suppose we have the sequence $a_0,a_1,a_2,\dots$ of positive reals. Suppose ftsoc that $a_{n-1}\geq 2^{-1/n}+2^{-1/n}a_n$ were true for all $n\geq 1$. Then note that $a_{n-1}>2^{-1/n}$. Suppose for $k\geq 0$ we have that $$a_{n-1}>2^{-1/n}+2^{-1/n-1/(n+1)}+\dots+2^{-1/n-1/(n+1)-\dots-1/(n+k)} \qquad (\star)$$for all $n$. Then $$a_{n-1}\geq 2^{-1/n}+2^{-1/n}a_n>2^{-1/n}+2^{-1/n}(2^{-1/(n+1)}+2^{-1/(n+1)-1/(n+2)}+\dots+2^{-1/(n+1)-1/(n+2)-\dots-1/(n+k+1)})=2^{-1/n}+2^{-1/n-1/(n+1)}+\dots+2^{-1/n-1/(n+1)-\dots-1/(n+k+1)}$$Thus, by induction, $(\star)$ in fact holds for all $k$. So, using the lemma we have that $$a_0>\sum_{m=1}^k 2^{-H_m}>\sum_{m=5}^k \frac 1m.$$Now, as $k\to \infty$, the sum on the RHS diverges which is absurd since $a_0$ is a finite number. Thus we have reached a contradiction. And so there is a number $p$ such that $1+a_p>a_{p-1}2^{1/p}$. Now define a new sequence $\{b_n\}$ such that $b_0=a_{p+1}, b_1=a_{p+2},\dots$. Applying the same method to this new sequence we will find a number $p'$ such that $1+b_{p'}>b_{p'-1}2^{1/p'}>b_{p'-1}2^{1/(p+p'+1)}$ and so $1+a_{p+p'+1}>a_{p+p'}2^{1/(p+p'+1)}$. Continuing like this we will find infinitely many $a_n$ satisfying the inequality.

We begin by showing that $\sqrt[n]{2} \leq \frac{n+1}{n}$. Let $\sqrt[n]{2}=1+a$. Notice that $1+an \leq \sum\limits_{i=0}^n \binom{n}{i}a^i=(1+a)^n=2$ which implies $a \leq \frac{1}{n}$. It immediately follows that $\sqrt[n]{2} = 1 + a \leq \frac{n + 1}{n}$. $\square$ It simply suffices to show that there always exists a positive integer satisfying $1 + a_n > a_{n-1}\sqrt[n]{2}$. Suppose that this is not true. Define the sequence $b_1, b_2, \dots$ by $b_0 = a_0$ and the relation $1+b_n=b_{n-1}\sqrt[n]{2}$. We show by induction that $a_n \leq b_n$. The base case $n = 1$ is trivial. Suppose that it holds for $n-1$. Then, we have \[a_n=a_{n-1} \sqrt[n]{2}-1 \leq b_{n-1}\sqrt[n]{2}-1 = b_n \textrm{.} \ \square\]Now, we show that $b$ will eventually become negative. It is clear that this will finish our problem. We begin by showing, using induction, that $b_n \leq b_0(n-1)-\sum\limits_{i=2}^n\frac{n+1}{i}-1$. The base case $n=1$ is trivial. Assume that it is true for $n-1$. Then, we have \[b_n = 1+b_{n-1}\sqrt[n]{2} \leq (b_0(n-2)-\sum\limits_{i=2}^{n-1}\frac{n}{i}-1)\frac{n+1}{n}-1=b_0(n-1)-\sum\limits_{i=2}^n\frac{n+1}{i}-1 \textrm{.} \ \square\]Recall that $\sum\limits_{i=1}^{\infty}\frac{1}{i}=\infty$. From this, we can infer that there exists a positive integer $k$ such that $\sum\limits_{i=2}^k\frac{1}{i} \geq b_0$. Then, $k$ satisfies $b_k < b_0(k+1)-\sum\limits_{i=2}^{k} \frac{k+1}{i} = (k+1)(b_0-\sum\limits_{i=2}^k \frac{1}{i}) \leq 0$, as desired. $\blacksquare$

By Bernoulli's inequality, it suffices to show that the inequality $$ 1 + a_n > a_{n-1} \left( \frac{n+1}{n} \right) $$holds for infinitely many positive integers $n$. To do this, assume the contrary, i.e., there exists $N \in \mathbb{Z}^+$ such that $$ 1 + a_n \leq a_{n-1} \left( \frac{n+1}{n} \right) $$for all integers $n \geq N$. We can show by recursion that $$ 1 + a_{N+m} \leq a_{N-1} \left( \frac{N+m+1}{N} \right) - \sum_{k=1}^{n+m} \left( \frac{N+m+1}{N+k} \right) $$where $$ \sum_{k=1}^{n+m} \left( \frac{1}{N+k} \right) \leq \frac{a_{N-1}}{N}. $$But this is not possible, since $$ \lim_{m \to \infty} \sum_{k=1}^{n+m} \left( \frac{1}{N+k} \right) = \infty $$and we are done.

Nice problem! Assume for the sake of contradiction that $1 + a_n \le a_{n-1} \cdot 2^{1/n}$ for all $n > N$ where $N$ is finite. Then we get $$a_n \le a_{n-1} \cdot 2^{1/n} - 1.$$ Claim: For all $k \in \mathbb{N}$ we have $$a_{N+k} \le a_{N} 2^{\frac{1}{N+1} + \dots + \frac{1}{N+k}} - 2^{\frac{1}{N+2} + \dots + \frac{1}{N+k}} - 2^{\frac{1}{N+3} + \dots + \frac{1}{N+k}} - \dots - 2^{\frac{1}{N+k}} - 1.$$Proof: The proof is by induction on $n$ with the base case $k=1$ being by our initial assumption. Now assume that the inductive hypothesis holds for $k$; we will show it holds for $k+1.$ We know that $a_{N+k+1} \le a_{N+k} \cdot 2^{\frac{1}{N+k+1}} - 1.$ Plugging our inductive hypothesis assertion in and distributing the $2^{\frac{1}{N+k+1}}$ factor completes our induction. Now, the positive real numbers condition implies that $a_{N+k} > 0$ for all $k.$ Using our claim, this gives $$a_N \ge 2^{-\frac{1}{N+1}} + 2^{-\frac{1}{N+1} - \frac{1}{N+2}} + \dots + 2^{-\frac{1}{N+1} - \frac{1}{N+2} - \dots - \frac{1}{N+k}}$$for all $k.$ However: Claim: The sum $$2^{-H_1} + 2^{-H_2} + \dots$$diverges, where the $H_k$ are the harmonic numbers. Proof: A lower rectangle approximation on the curve $y = \frac{1}{x}$ yields $H_i < \ln(k) + 1$ so that $$2^{-H_k} > \frac{1}{2} \cdot 2^{-\ln(k)} = \frac{1}{2} \cdot k^{-\ln(2)}.$$Since $\ln(2) < 1,$ summing gives a sum greater than the harmonic series which diverges, as claimed. By some simple manipulations, we get that the series $$2^{-\frac{1}{N+1}} + 2^{-\frac{1}{N+1} - \frac{1}{N+2}} + \dots$$diverges, so $a_N$ is infinite, contradiction.