All Questions
8
questions
5
votes
1
answer
205
views
Prove $\sum_{n=1}^{\infty}\frac{\Gamma(n+\frac{1}{2})}{(2n+1)(2n+2)(n-1)!}=\frac{(4-π)\sqrt{\pi}}{4}$
Prove
$$S=\sum_{n=1}^{\infty}\frac{\Gamma(n+\frac{1}{2})}{(2n+1)(2n+2)(n-1)!}=\frac{(4-π)\sqrt{\pi}}{4}.$$
I don't know how to evaluate this problem .
At first I used partial fraction but I got ...
- 751
0
votes
0
answers
196
views
Closed form for this sum involving the lower incomplete gamma function?
Can this sum be written in simpler terms?
$$\sum_{k=0}^\infty \frac{1}{z-k} \cdot \frac{\gamma(k,-\log x)}{\Gamma(k)}$$
(where $\gamma(k,-\log x)$ is the lower incomplete gamma function)
I'm pretty ...
- 987
0
votes
1
answer
228
views
Infinite sum over Gamma functions?
I am having quite a bit of trouble understanding this sum.
Can someone explain to me exactly how to this from 1 to 3,very easily way?
Question its from this webpage
Thanks.
- 4,492
3
votes
2
answers
807
views
Closed form for an infinite sum over Gamma functions?
I am having quite a bit of trouble trying to find a closed form (or a really fast way to compute) for the infinite sum
$$\sum_{n=1}^{\infty} a^n \dfrac{\gamma(n+1,b)}{\Gamma(n+1)\Gamma(n)}$$
where $\...
- 33
2
votes
0
answers
238
views
Closed form expression for a sum
I want to calculate a sum of the form $$\sum_{k=0}^m \frac{\Gamma[m+1+\alpha-k]^2}{\Gamma[m+1-k]^2}\frac{\Gamma[x+k]}{\Gamma[x]k!}$$ where $m>0$ and belongs to integers and $\alpha$ takes half ...
- 21
5
votes
3
answers
1k
views
Hypergeometric 2F1 with negative c
I've got this hypergeometric series
$_2F_1 \left[ \begin{array}{ll}
a &-n \\
-a-n+1 &
\end{array} ; 1\right]$
where $a,n>0$ and $a,n\in \mathbb{N}$
The problem is that $-a-n+1$ is ...
- 159
18
votes
1
answer
400
views
Closed form for $\sum_{n=1}^\infty\frac{(-1)^n n^a H_n}{2^n}$
Is there a closed form for the sum $$\sum_{n=1}^\infty\frac{(-1)^n n^a H_n}{2^n},$$ where $H_n$ are harmonic numbers: $$H_n=\sum_{k=1}^n\frac{1}{k}=\frac{\Gamma'(n+1)}{n!}+\gamma.$$
This is a ...
- 2,575
26
votes
3
answers
1k
views
Closed form for $\sum_{n=1}^\infty\frac{(-1)^n n^4 H_n}{2^n}$
Please help me to find a closed form for the sum $$\sum_{n=1}^\infty\frac{(-1)^n n^4 H_n}{2^n},$$ where $H_n$ are harmonic numbers: $$H_n=\sum_{k=1}^n\frac{1}{k}=\frac{\Gamma'(n+1)}{n!}+\gamma.$$
- 2,575