4
$\begingroup$

$$\left(\frac{1}{3}\right)^{2}+\left(\frac{1\cdot2}{3\cdot5}\right)^{2}+\left(\frac{1\cdot2\cdot3}{3\cdot5\cdot7}\right)^{2}+...$$

I am not able to find a general equation and that's creating problem for me as I can't proceed further without it.

$\endgroup$
  • $\begingroup$ If you want a formula for the sum in well-known terms, then you can certainly use some Hypergeometric function. But it is not clear to me if that is what you are after. $\endgroup$ – mickep Sep 2 '17 at 15:47
  • $\begingroup$ @Did are you sure? when n=1 it doesn't hold so. $\endgroup$ – Iti Shree Sep 2 '17 at 15:52
  • $\begingroup$ @ItiShree Good catch, I missed some powers of $2$ (shame on me). Anyway, you have simpler approaches below now, so all is well. $\endgroup$ – Did Sep 2 '17 at 15:54
  • $\begingroup$ @Did It's fine. $\endgroup$ – Iti Shree Sep 2 '17 at 16:45
  • $\begingroup$ it is extremely bad form not to have the general rule for the terms, it makes it ambiguous and unclear, go shame your professor! also if you want you can make as many general equations having the given first 3 terms. $\endgroup$ – Arjang Dec 16 '17 at 5:59
7
$\begingroup$

The generic term of the series is $$a_n=\prod_{k=1}^n \left(\frac{k}{2k+1}\right)^2.$$ Hence, as $n\to+\infty$, $$\frac{a_{n+1}}{a_n}= \left(\frac{(n+1)}{2(n+1)+1}\right)^2\to \frac{1}{4}.$$ What may we conclude?

$\endgroup$
  • $\begingroup$ I was hoping to see this solution. My profs just started this chapter and they didn't mentioned we can use $\pi$ that's why I had some doubt. But thanks. $\endgroup$ – Iti Shree Sep 2 '17 at 15:53
  • $\begingroup$ @ItiShree If by $\pi$ you mean $\prod$ then that is just notation, use it or not you get the same answer. $\endgroup$ – user8277998 Sep 2 '17 at 16:20
2
$\begingroup$

Yes it converges of course! Use $$\frac{n}{2n+1}<\frac{1}{2}$$

We obtain: $$\left(\frac{1}{3}\right)^{2}+\left(\frac{1.2}{3.5}\right)^{2}+\left(\frac{1.2.3}{3.5.7}\right)^{2}+...<\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+...=\frac{\frac{1}{4}}{1-\frac{1}{4}}=\frac{1}{3}.$$

$\endgroup$
  • $\begingroup$ The problem is the series can't be converted in a formula then how can I proceed for the summation? shall I just use comparison test with $\sum_{n=1}^{\infty} (\frac{1}{2})^n$? $\endgroup$ – Iti Shree Sep 2 '17 at 15:42
  • $\begingroup$ @Iti Shree I added something. See now please. $\endgroup$ – Michael Rozenberg Sep 2 '17 at 15:50
  • $\begingroup$ I understand now, thanks. $\endgroup$ – Iti Shree Sep 2 '17 at 15:51
  • $\begingroup$ @Iti Shree You are welcome! $\endgroup$ – Michael Rozenberg Sep 2 '17 at 15:51
2
$\begingroup$

We have to deal with: $$ \sum_{n\geq 1}\left(\frac{n!}{(2n+1)!!}\right)^2=\sum_{n\geq 1}\left(\frac{2^n n!^2}{(2n+1)!}\right)^2 = \sum_{n\geq 1}\frac{4^n}{(2n+1)^2\binom{2n}{n}^2} $$ and since $\binom{2n}{n}\sim\frac{4^n}{\sqrt{\pi n}}$, the above series is clearly convergent. We may recall that $$ \frac{\arcsin z}{\sqrt{1-z^2}}=\sum_{n\geq 0}\frac{z^{2n+1}4^n}{(2n+1)\binom{2n}{n}} \tag{1}$$ $$ \int_{0}^{\pi/2}\left(\sin\theta\right)^{2n+1}\,d\theta = \frac{4^n}{(2n+1)\binom{2n}{n}}\tag{2} $$ hence we have: $$ \sum_{n\geq 1}\left(\frac{n!}{(2n+1)!!}\right)^2=\tfrac{1}{9}\cdot\phantom{}_3 F_2\left(1,2,2;\tfrac{5}{2},\tfrac{5}{2};\tfrac{1}{4}\right)=-1+\int_{0}^{\pi/2}\frac{4\arcsin\frac{\sin\theta}{2}}{\sqrt{4-\sin^2\theta}}\,d\theta.\tag{3} $$ or, in a more compact form: $$ \sum_{n\geq 1}\left(\frac{n!}{(2n+1)!!}\right)^2= -1+\int_{0}^{\pi/6}\frac{4\theta\,d\theta}{\sqrt{1-4\sin^2\theta}}.\tag{4}$$

$\endgroup$
  • $\begingroup$ +1 you are the guy who prefers not to use any convergence tests, but rather show that the series is convergent by finding its sum explicitly. $\endgroup$ – Paramanand Singh Dec 16 '17 at 5:33

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.