0
$\begingroup$

I need to prove that the sum of the following series is convergent to : $1 \ge Sum$

$$\sum_{n=1}^\infty \ \left[\left(\frac{2n+1}{n}\right)\left(\frac{2n+2}{n}\right)\cdots\left(\frac{2n+n}{n}\right)\right]^{-1}$$

Well I succeed to compare it to another series but could only prove its $2 \ge Sum$.

I would like to get a hint and explanation about the rationale of the solution.

Thanks!

$\endgroup$
  • 1
    $\begingroup$ Each of the terms $\frac{2n+k}{n}$ is $>2$, so we have $\sum_{n=1}^\infty a_n$ where $a_n<\frac{1}{2^n}$. $\endgroup$ – almagest Apr 23 '16 at 12:04
  • $\begingroup$ @OlivierOloa I think there is a tiny $]^{-1}$ at the end! $\endgroup$ – almagest Apr 23 '16 at 12:06
  • $\begingroup$ @almages That's true .. watch out from the $]^{-1}$ $\endgroup$ – Barak Mi Apr 23 '16 at 12:07
  • $\begingroup$ And you should know that $\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\dots=1$ $\endgroup$ – almagest Apr 23 '16 at 12:09
  • $\begingroup$ The dummy variable is $i$ or $n$? $\endgroup$ – Olivier Oloa Apr 23 '16 at 12:11
0
$\begingroup$

Since $(2n+1)\ldots (2n+n)\geq (2n)^n=2^n n^n$,

$\displaystyle \sum_{n=1}^N \frac{n}{2n+1}\ldots\frac{n}{2n+n} \leq \sum_{n=1}^N \frac{1}{2^n} \leq \sum_{n=1}^\infty \frac{1} {2^n}=1$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.