All Questions
46
questions
42
votes
3
answers
1k
views
Calculate the following infinite sum in a closed form $\sum_{n=1}^\infty(n\ \text{arccot}\ n-1)$?
Is it possible to calculate the following infinite sum in a closed form? If yes, please point me to the right direction.
$$\sum_{n=1}^\infty(n\ \text{arccot}\ n-1)$$
- 13.2k
15
votes
2
answers
865
views
Evaluate $\sum\limits_{n=1}^{\infty}\frac{1}{n^3}\binom{2n}{n}^{-1}$. [duplicate]
Evaluate $$\sum\limits_{n=1}^{\infty}\frac{1}{n^3}\binom{2n}{n}^{-1}.$$
My work so far and background to the problem.
This question was inspired by the second page of this paper. The author of the ...
- 6,560
14
votes
3
answers
2k
views
Closed form for $\sum_{n=1}^{\infty}\frac{1}{\sinh^2\!\pi n}$ conjectured
By trial and error I have found numerically
$$\sum_{n=1}^{\infty}\frac{1}{\sinh^2\!\pi n}=\frac{1}{6}-\frac{1}{2\pi}$$
How can this result be derived analytically?
- 2,903
14
votes
2
answers
482
views
Closed-form of $\sum_{n=0}^\infty\;(-1)^n \frac{\left(2-\sqrt{3}\right)^{2n+1}}{(2n+1)^2\quad}$
The following question is purely my curiosity. During my calculation to answer @Chris'ssis's question in chat room I encountered this series
$$\sum_{n=0}^\infty\; \frac{\left(2-\sqrt{3}\right)^{2n+1}}{...
- 11k
11
votes
2
answers
689
views
Infinite Series $\sum\limits_{k=1}^{\infty}\frac{k^n}{k!}$
How can I find the value of the sum $\sum_{k=1}^{\infty}\frac{k^n}{k!}$?
for example for $n=6$, we have
$$\sum_{k=1}^{\infty}\frac{k^6}{k!}=203e.$$
- 5,626
10
votes
3
answers
190
views
Show that $\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^2} \sum_{n=1}^k \frac{1}{n}=\frac{5\zeta(3)}{8}$
$$\sum_{k=1}^{\infty} \dfrac{(-1)^{k+1}}{k^2} \sum_{n=1}^k \dfrac{1}{n}=\frac{5\zeta(3)}{8}$$
I tried to create a proof from some lemmas some are suggested by my Senior friends
Lemma 1 $$
{H_n} = \sum\...
user1254447
8
votes
6
answers
325
views
Calculate the closed form of the following series
$$\sum_{m=r}^{\infty}\binom{m-1}{r-1}\frac{1}{4^m}$$
The answer given is $$\frac{1}{3^r}$$ I tried expanding the expression so it becomes $$\sum_{m=r}^{\infty}\frac{(m-1)!}{(r-1)!(m-r)!}\frac{1}{4^m}$$...
user879314
7
votes
6
answers
242
views
Find triple summation rel in a closed form $S=\sum_{n=1}^{\infty}\sum_{m=1}^{n}\sum_{k=1}^{m}\frac{1}{(n+1)(k+1)(m+1)nmk}$
Evaluate $\displaystyle S=\sum_{n=1}^{\infty}\sum_{m=1}^{n}\sum_{k=1}^{m}\frac{1}{(n+1)(k+1)(m+1)nmk}$
My attempt :
Let $$A=\sum_{k=1}^{m}\frac{1}{k(k+1)}
=\sum_{k=1}^{m}\left( \frac1{k}-\frac1{...
- 2,323
6
votes
5
answers
512
views
Infinite Series $1+\frac12-\frac23+\frac14+\frac15-\frac26+\cdots$
Was given the following infinite sum in class as a question, while we were talking about Taylor series expansions of $\ln(1+x)$ and $\arctan(x)$:
$$1+\frac12-\frac23+\frac14+\frac15-\frac26+\cdots$$
...
- 61
6
votes
2
answers
242
views
How to prove $\sum_{n=0}^{\infty} \frac {(2n+1)!} {2^{3n} \; (n!)^2} = 2\sqrt{2} \;$?
I found out that the sum
$$\sum_{n=0}^{\infty} \frac {(2n+1)!} {2^{3n} \; (n!)^2}$$
converges to $2\sqrt{2}$.
But right now I don't have enough time to figure out how to solve this.
I would ...
- 63
6
votes
4
answers
241
views
Evaluate $\int_{0}^{1}\{1/x\}^2\,dx$
Evaluate
$$\displaystyle{\int_{0}^{1}\{1/x\}^2\,dx}$$
Where {•} is fractional part
My work
$$\displaystyle{\int\limits_0^1 {{{\left\{ {\frac{1}{x}} \right\}}^2}dx} = \sum\limits_{n = 1}^\infty {\...
user1235604
6
votes
2
answers
288
views
Closed-form of $\sum_{k=0}^{\infty} \frac{k^a\,b^k}{k!}$
While working on this question I think I've found a closed-form expression for the following series, but I don't know how to prove it.
Let $a \in \mathbb{N}$ and $b \in \mathbb{R}$. Then
$$\sum_{k=0}...
- 12.4k
6
votes
3
answers
169
views
evaluate the summation : $\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(n+1)(n+2x+3)}$
Find
$$S=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(n+1)(n+2x+3)}$$
for $x≥0$.
At first, I use a partial fraction
$$S=\displaystyle\sum_{n=0}^{\infty}\left(\frac{(-1)^{n}}{2(x+1)(n+1)}-\frac{(-...
- 2,323
6
votes
2
answers
185
views
Calculate $\sum\limits_{n = - \infty }^\infty {\frac{{\log \left( {{{\left( {n + \frac{1}{3}} \right)}^2}} \right)}}{{n + \frac{1}{3}}}} $
question:
how do we find that:
$$ S = \sum\limits_{n = - \infty }^\infty {\frac{{\log \left( {{{\left( {n + \frac{1}{3}} \right)}^2}} \right)}}{{n + \frac{1}{3}}}} $$
I modified the sum
$$\sum\...
user1254447
5
votes
2
answers
200
views
What is the close form of: $\sum\limits_{k=1}^{\infty}\log\left(\frac{1}{k^2}+1\right)$
Is there a close form for of this series
$$\sum_{k=1}^{\infty}\log\left(\frac{1}{k^2}+1\right) =\log \prod_{k=1}^{\infty}\left(\frac{1}{k^2}+1\right)$$
I know it converges in fact since $ \log(x+1)\...
- 24.2k