All Questions
15
questions
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
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
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
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
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
$$\sum_{n=0}^\infty\; \frac{\left(2-\sqrt{3}\right)^{2n+1}}{...
- 11k
8
votes
6
answers
325
views
Calculate the closed form of the following series
$$\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}$$...
user879314
6
votes
2
answers
242
views
How to prove $\sum_{n=0}^{\infty} \frac {(2n+1)!} {2^{3n} \; (n!)^2} = 2\sqrt{2} \;$?
I found out that the sum
$$\sum_{n=0}^{\infty} \frac {(2n+1)!} {2^{3n} \; (n!)^2}$$
converges to $2\sqrt{2}$.
But right now I don't have enough time to figure out how to solve this.
I would ...
- 63
6
votes
2
answers
288
views
Closed-form of $\sum_{k=0}^{\infty} \frac{k^a\,b^k}{k!}$
While working on this question I think I've found a closed-form expression for the following series, but I don't know how to prove it.
Let $a \in \mathbb{N}$ and $b \in \mathbb{R}$. Then
$$\sum_{k=0}...
- 12.4k
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
3
votes
5
answers
390
views
Infinite Series $\left(\frac12+\frac14-\frac23\right)+\left(\frac15+\frac17-\frac26\right)+\left(\frac18+\frac{1}{10}-\frac29\right)+\cdots$ [duplicate]
How do I find the sum of the following infinite series:
$$\left(\frac12+\frac14-\frac23\right)+\left(\frac15+\frac17-\frac26\right)+\left(\frac18+\frac{1}{10}-\frac29\right)+\cdots$$
The series ...
- 375
1
vote
5
answers
5k
views
Find a formula for $\sum_{i=1}^n (2i-1)^2 = 1^2+3^2+....+(2n-1)^2$
Consider the sum
$$\sum_{i=1}^n (2i-1)^2 = 1^2+3^2+...+(2n-1)^2.$$
I want to find a closed formula for this sum, however I'm not sure how to do this. I don't mind if you don't give me the answer but ...
- 11
0
votes
1
answer
189
views
How to solve this summation (Lerch Transcendent)?
How is it possible to deduce the closed form of the following?
$$\sum_{i = 0}^{n - 1} \frac{2^i}{n - i} = ?$$
0
votes
1
answer
72
views
some identity of summation and generalization
I see this way and idea from Simply Beautiful Art
profile here
Let : $S=\displaystyle\sum_{1≤k≤m≤n≤\infty}f(m)f(k)f(n)$
then :
$6S=\displaystyle\sum_{n,m,k≥1}f(m)f(k)f(n)+3\displaystyle\sum_{n,m≥...
- 2,323
-1
votes
2
answers
96
views
Evaluate $\sum_{m=1}^{\infty} \frac{(-1)^m \mathcal{H}_{2m} }{2m+1} - \frac{1}{2} \sum_ {m=1}^{\infty} \frac{(-1)^m \mathcal{H}_m}{2m+1}$ [duplicate]
Let's declare $\mathcal{G}$ is constant of Catalanand the $\mathcal{H}_m-st$ mharmonic term. Let it be shown that:
$$\displaystyle{\sum_{m=1}^{\infty} \frac{(-1)^m \mathcal{H}_{2m} }{2m+1} -\frac{1}{2}...
