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I was having trouble with the sum
$$\sum^{\infty}_{n=1} \frac{1}{n {2n\choose n} }$$

My Attempt $$S=\sum^{\infty}_{n=1} \frac{1}{n {2n\choose n} }=\sum^{\infty}_{n=1} \frac{n!\; n!}{n\;(2n)! }=\sum^{\infty}_{n=1} \frac{n(n-1)!\; n(n-1)!}{2n^2\;(2n-1)! } = \frac 1 2 \sum^{\infty}_{n=1}\frac{\Gamma(n)^2}{\Gamma(2n)}$$

$$S=\sum^{\infty}_{n=1}\frac{\Big(\int_0^{\infty}x^{n-1}e^{-x}dx\Big)^2}{\int_0^{\infty}x^{2n-1}e^{-x}dx}$$

What am I doing wrong? How do I proceed? How do I solve this integral?

the_candyman
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hwood87
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    Does this post help you? https://math.stackexchange.com/questions/1870690/evaluating-the-sums-sum-limits-n-1-infty-frac1n-binomknn-with-k – dust05 Sep 06 '21 at 06:46

1 Answers1

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Use $${n\choose k}^{-1}=(n+1) \int_{0}^{1} x^k(1-x)^{n-k} dx.$$ Then $$S=\sum_{n=1}^{\infty} \frac{1}{n{2n \choose n}}= \int_{0}^{1} dx ~\sum_{n=1}^{\infty}\frac{2n+1}{n} (x(1-x))^n=\int_{0}^{1} \frac{2x(1-x)}{1-x+x^2}dx-\int_{0}^{1}\log(1-x+x^2) dx =$$ $$\implies S=-2+2 \int_{0}^{1} \frac{dx}{1-x+x^2} -x \log(1-x+x^2)|_{0}^{1}+\int_{0}^{1} \frac{x(2x-1)dx}{1-x+x^2}.$$ Next, these integrals can be done easily to get the result.

Z Ahmed
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  • Why does $dx$ feature twice in your integral on the last line? And how is it that you were able to write $\int_0^1 \ln(1-x+x^2),dx$ as $x\ln(1-x+x^2)\Big|_0^1$? – FShrike Sep 06 '21 at 09:34
  • Corrected the typo od $dx$, Do it by parts taking 1 as second and $\log(1-x+x^2)$ as the first function. The last integrand has been corrected. – Z Ahmed Sep 06 '21 at 09:42