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When $a_k=\frac{k^2}{k^2+1}$

$b_n=a_1a_2a_3...a_n$

What I need to prove is $\lim_{n\to\infty}b_n\gt0$.

I think this limit is$\frac1 {10}$.

Similar problem is $a_k=\frac {k^2}{k^2+1}$

and I used $a_k \gt \frac {k^2-1}{k^2}$

so $b_n \gt \frac{n+1}{4n}$. then $\lim{b_n}\gt 0$

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sorry for my mistake

I want to prove when $a_k=\frac{k^2}{k^2+\sqrt{k}}$

Shootforthemoon
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Hwan
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1 Answers1

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More generally, we consider the sequence $\langle a_n\rangle$ with $a_n=\frac{k^2}{k^2+k^\alpha}$, for some $\alpha<1$.

Applying the well-known inequality $1+x\leq e^x$ with $x=1/k^{2-\alpha}$, we obtain $$a_k\geq e^{-1/k^{2-\alpha}}\quad\hbox{ for $k\geq1$},$$ Thus, for $n\geq1$, we have $$ b_n\geq \exp\left(-\sum_{k=1}^n\frac{1}{k^{2-\alpha}}\right) >\exp(-\zeta(2-\alpha))>0 $$ This proves the desired conclusion for any such sequence $\langle a_n\rangle$.

Mathematica yields $0.108686$ as an approximation for $\lim\limits_{n\to\infty}b_n$ when $\alpha=1/2$.

Omran Kouba
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  • I sorry my mistake. i wnat to show when $a_k=\frac {k^2}{k^2+\sqrt{k}}$ – Hwan Apr 07 '14 at 17:25
  • The principle used covers also that case, you have $a_k > 1 - \frac{1}{k^{3/2}}$ then, and $\sum k^{-3/2} < \infty$ too. Of course the bounds for the product change. – Daniel Fischer Apr 07 '14 at 17:34
  • (+1) Good answer. I know how frustrating it is to write an answer to a question that gets changed. – robjohn Apr 07 '14 at 17:34
  • @DanielFischer: unfortunately, the nice telescoping sum needs to be changed. – robjohn Apr 07 '14 at 17:36
  • Ugh, I wrote up pretty much the same thing. I didn't see your edit until I was done. Good answer again! – robjohn Apr 07 '14 at 18:02