Hi, I think I've got a solution for part a). The interesting is that I actually did not need $a_0 \leq a_1 \leq ...$, only that $a_0=1$ and $a_k > 0$ I proved that for all $n \geq 1:$ $b_n \leq 2-2\cdot\frac{1}{\sqrt{a_n}}$ holds, which is a very simple induction proof..... As to part b)...I'm still working on it....

Yimin Ge wrote: Hi, I think I've got a solution for part a). The interesting is that I actually did not need $a_0 \leq a_1 \leq ...$, only that $a_0=1$ and $a_k > 0$ I proved that for all $n \geq 1:$ $b_n \leq 2-2\cdot\frac{1}{\sqrt{a_n}}$ holds, which is a very simple induction proof..... As to part b)...I'm still working on it.... sorry,I’m not very sure about your solution for part a)...I think the $a_0 \leq a_1 \leq ...$ do need. For part b):juat take $a_n=x^n$ let $x \to 1^{+}$

Well, according to my gut feeling, it is possible because if you take a number smaller than the previous one, you have a negative summand and therfore you somehow "ease" the inequality (though the following summand will be greater but this might cancel) but well, I'm talking to mathematicians and not to physicians so I'll have to prove my claim I'm going to prove that $b_n\leq 2-2\cdot\frac{1}{\sqrt{a_n}}$ by induction. Let $u_n = \sqrt{a_n}$ for all $n\geq 0$ (just because I dont want to write so many $\sqrt{\ }$. $\Rightarrow b_n = \sum_{k=1}^n(1-\frac{u_{k-1}^2}{u_k^2})\frac{1}{u_k}$ i.b.: For $n=1$ we have $b_1=(1-\frac{1}{u_1^2})\frac{1}{u_1} \leq 2-2\frac{1}{u_1}$ which is after some standard transforming equivalent to $0\leq 2u_1^3-3u_1^2+1=(u_1-1)^2(2u_1+1)$ Now we deduce from n to n+1 i.p.: $b_n=\sum_{k=1}^n(1-\frac{u_{k-1}^2}{u_k^2})\frac{1}{u_k} \leq 2-2\frac{1}{u_n}$ i.c.: $b_{n+1}=\sum_{k=1}^{n+1}(1-\frac{u_{k-1}^2}{u_k^2})\frac{1}{u_k} \leq 2-2\frac{1}{u_{n+1}}$ Using the i.p. we obtain $\sum_{k=1}^{n+1}(1-\frac{u_{k-1}^2}{u_k^2})\frac{1}{u_k} \leq 2-2\frac{1}{u_n} + (1-\frac{u_n^2}{u_{n+1}^2})\frac{1}{u_{n+1}}$ so it remeans to be shown that $2-2\frac{1}{u_n} + (1-\frac{u_n^2}{u_{n+1}^2})\frac{1}{u_{n+1}} \leq 2-2\frac{1}{u_{n+1}}$. After some simplifying and some standard transforming we obtain $0\leq 2u_{n+1}^3-3u_{n+1}^2u_n + u_n^3 = (u_{n+1}-u_n)^2(2u_{n+1}+u_n)$. Assuming that I did not miscalculate (which would be very embarrassing though) $u_{n+1} \geq u_n$ is not necessary as we have $(u_{n+1}-u_n)^2$.....

yes,you are right.$1=a_0\le a_1\le a_2\le\ldots$ for part a) and b) just use $a_i \ge 0 (i=1,2,\ldots)$ and this problem is very strange,for part b) we must fix $x$ first ,then fix $n$. you will easily see $\lim_{n \to \infty}b_n=1+ \sqrt{x}$

Alternative proof for (a): Claim: $\frac{1 - \frac{a_{k - 1}}{a_{k}}}{\sqrt{a_{k}}} \leqslant 2\left(\frac{1}{\sqrt{a_{k - 1}}} - \frac{1}{\sqrt{a_{k}}}\right)$. Proof: Applying AM-GM to $\frac{a_{k - 1}}{a_{k}^{\frac{3}{2}}}, \frac{1}{\sqrt{a_{k - 1}}}, \frac{1}{\sqrt{a_{k - 1}}}$ gives \[ \frac{a_{k - 1}}{a_{k}^{\frac{3}{2}}} + \frac{2}{\sqrt{a_{k - 1}}} \geqslant \frac{3}{\sqrt{a_{k}}}, \]which rearranges to \[ \frac{1 - \frac{a_{k - 1}}{a_{k}}}{\sqrt{a_{k}}} = \frac{1}{\sqrt{a_{k}}} - \frac{a_{k - 1}}{a_{k}^{\frac{3}{2}}} \leqslant \frac{2}{\sqrt{a_{k - 1}}} - \frac{2}{\sqrt{a_{k}}} = 2\left(\frac{1}{\sqrt{a_{k - 1}}} - \frac{1}{\sqrt{a_{k}}}\right), \]as required. Thus $b_{n} = \sum_{k = 1}^{n} \frac{1 - \frac{a_{k - 1}}{a_{k}}}{\sqrt{a_{k}}} \leqslant \sum_{k = 1}^{n} 2\left(\frac{1}{\sqrt{a_{k - 1}}} - \frac{1}{\sqrt{a_{k}}}\right) = 2\left(\frac{1}{\sqrt{a_{0}}} - \frac{1}{\sqrt{a_{n}}}\right) < \frac{2}{\sqrt{a_{0}}} = 2$.

(b) Take $a_k=q^{-2k}$ where $0<q<1$. Then $b_n=q(1+q)(1-q^n)>2q^2(1-q^n)$. Take $n>\log_q(1-q)$. Then $b_n>2q^3$. Now take $q\ge\sqrt[3]{\frac{c}{2}}$ and gain $b_n>c$

frill, you could've simplified it: $\begin{aligned}\left(1-\dfrac{a_{k-1}}{a_k}\right)\cdot\frac1{\sqrt{a_k}}&=\dfrac{\sqrt{a_k}-\sqrt{a_{k-1}}}{\sqrt{a_k}}\cdot\dfrac{\sqrt{a_k}+\sqrt{a_{k-1}}}{\sqrt{a_k}}\cdot\dfrac1{\sqrt{a_k}}\\&\le\dfrac{\sqrt{a_k}-\sqrt{a_{k-1}}}{\sqrt{a_k}}\cdot\dfrac{2\sqrt{a_k}}{\sqrt{a_k}}\cdot\dfrac1{\sqrt{a_{k-1}}}\\&=2\dfrac{\sqrt{a_k}-\sqrt{a_{k-1}}}{\sqrt{a_k}\cdot\sqrt{a_{k-1}}}\\&=2\left(\frac1{\sqrt{a_{k-1}}}-\frac1{\sqrt{a_k}}\right)\end{aligned}$ However, this only proves $2$ is an upper bound, not $\sup\{x\in\Bbb R^+\mid x=b_n,n\in\Bbb N_0\}$. Now, we should find the optimal sequence so that $c=2-\varepsilon$ for some arbitrarily small $\varepsilon>0$ and $b_n\in(c,2)$ for sufficiently large $n\in\Bbb N$.