Sunday, December 4, 2016

sequences and series - Find the sum, if exists sumlimitsinftyn=1frac(2n)!22n(n!)2(n+1)


n=1(2n)!22n(n!)2(n+1)


By comparison test this series converges. Any nice way to work the sum?



I see that this can be written as: \sum\limits_{n=1}^{\infty}\dfrac{\binom{2n}{n}}{2^{2n}(n+1)}


Answer




Consider the following expansion: \frac{1}{\sqrt{1-4x}}=\sum_{n=0}^{\infty} {2n \choose n}x^n Integrate both the sides under the limit 0 to x i.e -\frac{1}{2}\left(\sqrt{1-4x}-1\right)=\sum_{n=0}^{\infty} {2n \choose n}\frac{x^{n+1}}{n+1} \Rightarrow -\frac{1}{2x}\left(\sqrt{1-4x}-1\right)=\sum_{n=0}^{\infty} {2n \choose n}\frac{x^{n}}{n+1} Substitute x=1/4 to get: \sum_{n=0}^{\infty} \frac{{2n \choose n}}{n+1}\frac{1}{4^n}=2 \Rightarrow \sum_{n=1}^{\infty} \frac{{2n \choose n}}{n+1}\frac{1}{4^n}=1


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