Fairly Impossible#2

Calculus Level pending

$\sum_{n=1}^{\infty} \frac{\Gamma(n)}{n\Gamma \left(n+\frac 12\right)}$

The above sum can be written as $\dfrac{\psi'(a)}{\Gamma(b)}$ , find $a+b$ .

Notation: $\psi'(\cdot)$ denotes derivative of digamma function .

The answer is 1.

1 solution

Aruna Yumlembam
Jun 21, 2020

Since,

$\sum_{n=1}^{\infty}\frac{1}{n}\Beta(s,n)$ $=$ $\psi^1(s)$ . Then saying $s=1/2$ we have ,

$\sum_{n=1}^{\infty}\frac{1}{n}\frac{\Gamma(n)}{\Gamma(1/2+n)}$ equal to $\frac{\psi^1(1/2)}{\Gamma(1/2)}$ .Hence $a=1/2,b=1/2$ ,then $a+b=1$ . Here $\Beta(s,n)=\frac{\Gamma(s)\Gamma(n)}{\Gamma(s+n)}$ .

For more "making sense" go to discussion name The series representation of Polygamma Function.Plus for people who have solved it on their own I would like to call them genius.

Aruna Yumlembam - 11 months, 3 weeks ago

Fairly Impossible#2

The answer is 1.

1 solution

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