Can we prove the equality: $$ \sum_{n\ =\ 1}^{\infty}\left(\, 1 \over 4\,\right)^{n}\, {\left(\, 2n\,\right)! \over \left(\, n!\,\right)^{2}} = 1 $$ or the equality doesn't hold actually ?.
Asked
Active
Viewed 90 times
1
-
2$\displaystyle\sum_{n=0}^\infty{2n\choose n}x^n=\frac1{\sqrt{1-4x}}.~$ See binomial series. – Lucian Dec 02 '14 at 02:32
-
Can you answer useing Gauss's Test: http://mathworld.wolfram.com/GausssTest.html – GEdgar Dec 02 '14 at 02:34
-
possible duplicate of Convergence of $\sum \frac{(2n)!}{n!n!}\frac{1}{4^n}$ – Dec 05 '14 at 04:50
-
@StevenTaschuk That question is about convergence of the series; this one is about the value of its sum. – Dec 05 '14 at 07:02
2 Answers
3
The sum of the first three terms is $$\frac12+\frac38+\frac{5}{16}=\frac{19}{16}\ ,$$ and all terms are positive. So the sum either converges to a value greater than $1$, or diverges.
David
- 82,662
-
-
Not surprised, but of course this is not needed in order to answer the OP's question. Lukas Geyer also showed that the series diverges, but had to use Stirling's approximation which is very hi-tech compared with my method ;-) – David Dec 04 '14 at 09:26
1
Using Stirling's formula, the terms in the sum are asymptotically $$ \frac{1}{4^n} \frac{\sqrt{4\pi n}\left(\frac{2n}{e}\right)^{2n}} {\left(\sqrt{2\pi n} \left(\frac{n}{e}\right)^n\right)^2} = \frac{\sqrt{4\pi n}}{2 \pi n} = \frac{1}{\sqrt{\pi n}} $$ (if my algebra is correct), which gives you a divergent series.
Lukas Geyer
- 18,259