# Finding the sum of $\sum_{n=1}^{\infty } \frac{4^{n}}{n\left ( 2n+1 \right )\binom{2n}{n}}$

Since $$\sum_{n=1}^{\infty } \frac{4^{n}}{n\left ( 2n+1 \right )\binom{2n}{n}} = \sum_{n=1}^{\infty } \frac{4^{n} \Gamma \left ( n \right )\Gamma \left ( n+1 \right )}{ \Gamma \left ( 2n+2 \right )} =\sum_{n=1}^{\infty } \frac{4^{n} \operatorname{B}\left ( n+1,n+1 \right )}{n},$$

then $$\sum_{n=1}^{\infty } \frac{4^{n} \operatorname{B}\left ( n+1,n+1 \right )}{n} =\int_{0}^{1}\sum_{n=1}^{\infty } \frac{4^{n}}{n} \left ( t(1-t) \right )^{n} dt.$$

Now, I've been stuck here for a day and cannot find this infinite sum.

Please give me some advice. Thank you in advance.

• The integral of the sum is the same as the sum of the integral, if that helps. – Toby Mak Feb 6 at 11:52

## 1 Answer

From the power series for $$\ln (1-x)$$, this is the easily evaluated$$-\int_0^1\ln (1-4t(1-t))dt=-2\int_0^1 \ln |1-2t|dt=-4\int_0^{1/2}\ln (1-2t)dt.$$

• what happen if we bound summation from 1 to 1000 ? – ABCDEFG user157844 Feb 6 at 12:19
• @ABCDEFGuser157844 I doubt there's a nice analytic treatment for very large but finite bounds. – J.G. Feb 6 at 13:08