http://www.mathlinks.ro/viewtopic.php?search_id=2008794938&t=152341

Can we say this? $ \sum_{m=1}^{\infty}lim_{n\rightarrow{\infty}}(b_{m}+b_{2m}+...+b_{nm})=\sum_{m=1}^{\infty}{d(m)b_{m}}=\sum_{m=1}^{\infty}\frac{1}{m}$ (absurd) where $ d(m)$ is the number of divisor of m

perfect_radio what do you think?It's possible.

I guess, I found something interesting on this problem. The idea is the following: Write the sum $ b_{m}+b_{2m}+b_{3m}+...=b_{m}+(b_{2m}+b_{4m}+...)+(b_{3m}+b_{6m}+...)+...-$ $ -(b_{6m}+b_{12m}+...)-(b_{8m}+b_{16m}+...)+...=b_{m}+\frac{1}{2m}+\frac{1}{3m}+\frac{1}{5m}+...-$ $ -\frac{1}{6m}-\frac{1}{10m}-\frac{1}{14m}...=\frac{1}{m}$, from this we can see that $ b_{m}=\frac{1}{m}(1-\sum_{i\in P}{\frac{1}{i}}+\sum_{i,j\in P, i\neq j}{\frac{1}{ij}}-\sum_{i,j,k \in P, i\neq j\neq k, i\neq k}\frac{1}{ijk}+...)$, where $ P$ is the set of all prime numbers. Clearly the sum in the brackets is independent of $ m$ and generally of anything. There are two options, either it does not exist or it is some constant let's say $ k$. In latter case we have that $ k(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...)=1$, which brings again to impossible result.

Yes. Exclusiom-Inclusion Principle kills it.

Sure, the solution of post 5 is correct? As there has been working with infinitely many infinite series, it seems there will be trouble with the order of executing that sum. Also post 3 seems like not concluding anything as $d(m)$ isn't bounded.

Quote: Determine whether there exists a sequence $a_1,a_2,\ldots$ of nonnegative reals such that \[ a_{n}+a_{2n}+a_{3n}+\cdots=\frac1n\]for every positive integer $n$. No. Suppose there did. Let $S_k=\{ n\in \mathbb{N}:p_1\mid n, \text{ or } p_2\mid n, \ldots, \text{ or } p_k\mid n\}$ where we order the primes $\{p_1<p_2<\cdots\}$. The idea is to ``interlace'' the sequences for $n=p_1,\ldots,p_k$. Using CRT and Inclusion-Exclusion: \[ \sum_{n\in S_k} a_n = \sum_{\substack{T\subset S_k \\ T\not=\emptyset}} (-1)^{|T|+1}\prod_{i\in T}\frac{1}{p_i} = 1-\prod_{i=1}^k\left(1-\frac{1}{p_i}\right) \]by expansion. Using the estimate $\log(1-\varepsilon)\le -0.01\varepsilon$ for $\varepsilon\in (0,1)$, we have \[ \log \prod_{i=1}^k\left(1-\frac{1}{p_i}\right) = \sum_{i=1}^k \log\left(1-\frac{1}{p_i}\right) \le \sum_{i=1}^k -0.01\cdot \frac{1}{p_i}=-0.01P_k\]where $P_k:=\sum_{i=1}^k 1/p_i$. Therefore, since $\log$ is an increasing function, we have \begin{align*} \prod_{i=1}^k\left(1-\frac{1}{p_i}\right) \le e^{-0.01P_k} &\implies \sum_{n\in S_k}a_n \ge 1-\frac{1}{\left(e^{0.01} \right)^{P_k}}. \end{align*}Hence \begin{align*} 1-a_1=\sum_{i\ge 2}a_i \ge \sum_{n\in S_k} a_n \ge 1-\frac{1}{\left(e^{0.01} \right)^{P_k}} \implies a_1 \le \frac{1}{\left(e^{0.01} \right)^{P_k}}. \end{align*}But it is well-known that $P_k$ grows arbitrarily large varying $k$, so this forces $a_1=0$. Now, fix an $\ell$. Let $b_n=\ell a_{\ell n}$ for each $n$. Then $(b_n)$ is also a valid sequence, since \[ b_n+b_{2n}+\cdots = \ell a_{\ell n}+\ell a_{2\ell n} + \cdots = \ell \cdot \frac{1}{\ell n} = \frac{1}{n}.\]Therefore, $b_1=0$, so $\ell a_\ell=0$, i.e. $a_\ell=0$. Therefore, the entire sequence is $0$, contradiction.

The answer is no. Assume for contradiction such a sequence exists, and let \(P_n\) denote the first \(n\) primes. By the Principle of Inclusion-Exclusion, \[\frac1n-a_n=\lim_{k\to\infty}\sum_{T\subset P_k}\frac{(-1)^{|T|}}{n\prod_{p\in T}p}=\frac1n\lim_{k\to\infty}\sum_{T\subset P_k}\frac{(-1)^{|T|}}{\prod_{p\in T}p}=\frac{1-a_1}n.\]Solving, \(a_n=\tfrac1na_1\) for all \(n\), so \[1=\sum_{k=1}^\infty a_k=\sum_{k=1}^\infty\frac{a_1}k=a_1\sum_{k=1}^\infty\frac1k.\]Since \(\sum_{k=1}^\infty a_k\) diverges, \(a_1=0\), which means that \(a_n=0\) for all \(n\). This yields the desired contradiction.

i think the idea of this problem related to Euler formula in Zeta Function

Indeed. Note that $a_1 + a_2 + \dots = 1$ and that $a_n \le \frac{1}{n}$. However, we also have that \[ \sum_{k}^{N} \mu(k) (a_k + a_{2k} + \dots) = \sum_{k}^N \frac{\mu(k)}{k} \]eventually contains all of $a_1$ through $a_N$ and thus has a limit of $1$. This contradicts the fact that $\lim_{s \to 1^+} \frac{1}{\zeta(s)} = \lim_{s \to 1^+} \sum_k \frac{\mu(k)}{k}$ approaches $0$ which implies that this has a limit of $0$.

Let $D_i$ be the set of all positive integers with exactly $i$ prime factors, not necessarily distinct. Observe that $$ a_n = 1 - \sum_{k=2}^{\infty} a_{kn} = 1 - \sum_{k=1}^{\infty} (-1)^{k+1} \left( \sum_{x \in D_k} \left( \sum_{i=1}^{\infty} a_{xkn} \right) \right) = \frac{a_1}{n}. $$$$ =\frac{1}{n}\left(1 - \sum_{k=1}^{\infty} (-1)^{k+1} \left( \sum_{x \in D_k} \left( \sum_{i=1}^{\infty} a_{xk} \right) \right)\right) = \frac{a_1}{n}. $$But this implies that $$ a_1 + a_2 + a_3 + \dots = a_1 + \frac{a_1}{2} + \frac{a_1}{3} + \dots \neq 1 $$which is a contradiction.