How to calculate
$$\sum_{n=0}^{\infty}\binom{2n}{n}\frac{2n}{2^{2n}(2n-1)}$$
?
I tried to use residues, generating function, combinatorics formulas, but unsuccessfully.
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Hint: $~\displaystyle\sum_{n=0}^{\infty}\binom{2n}{n}~x^n~=~\frac1{\sqrt{1-4x}}~.~$ Now ask yourself how you can employ differentiation
and integration so as to force an extra $~\dfrac{2n}{2n-1}~$ into the formula.
Lucian
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See binomial series for more information. – Lucian May 01 '15 at 09:14
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$\sum_{n=0}^{\infty} \binom{2n}{n} \frac{2n}{2n-1}x^{2n-1}=\int_{0}^{x}\frac{1}{y}\frac{d}{dy}\left(\frac{1}{\sqrt{1-4y^2}}\right)dy$. If $x=\frac{1}{2}$ then we have $\sum_{n=0}^{\infty} \binom{2n}{n} \frac{2n}{2n-1}\frac{1}{2^{2n}}=\infty$. – mikis May 01 '15 at 09:36
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@mikis: Yes, $~\displaystyle\sum_{n=0}^{\infty}\binom{2n}{n}~\frac{2n}{2n-1}~x^n~=~\frac{4x}{\sqrt{1-4x}}~,~$ which indeed diverges for $x=\dfrac14~.$ – Lucian May 01 '15 at 09:44
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There is a simple way for proving that your series is divergent. Since: $$ \frac{1}{4^n}\binom{2n}{n}=\frac{(2n-1)!!}{(2n)!!}=\prod_{k=1}^{n}\frac{2n-1}{2n}\tag{1}$$ it follows that: $$ (2n+1)\left(\frac{1}{4^n}\binom{2n}{n}\right)^2 = \prod_{k=1}^{n}\left(1-\frac{1}{4n^2}\right)\tag{2} $$ and since the product in the RHS of $(2)$ is convergent, we have that $\frac{1}{4^n}\binom{2n}{n}$ behaves like $\frac{1}{\sqrt{n}}$ for large $n$s, so your series is divergent by the $p$-test. Anyway, a related convergent series is presented in this question.
Jack D'Aurizio
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