All Questions
20
questions
3
votes
0
answers
162
views
Closed form for $\sum_{k=n}^{2n-1} k(-1)^k \sum_{i=k}^{2n-1} {2n \choose i+1} {i \choose n}$?
I've found this sum:
$$S(n) = \sum_{k=n}^{2n-1} k(-1)^k \sum_{i=k}^{2n-1} {2n \choose i+1} {i \choose n}$$
The inner sum is elliptical iirc, but perhaps the double sum has a nice expression. We can ...
2
votes
1
answer
139
views
Finding the value of $\sum^{\infty}_{n=3,5,7,9....}\frac{2n^2 \exp(-\pi n/2)}{\exp(\pi n)+1}$
I want to evalutate:
$\displaystyle \tag*{} \sum \limits ^{\infty}_{n=3,5,7,9....}\dfrac{2n^2 \exp \left(-\pi n/2\right)}{\exp(\pi n)+1}$
This question is inspired from my previous question which ...
2
votes
0
answers
28
views
Summation with given parameter :
Consider the following sum:
$$F(x)=\sum_{n=2}^{\infty}\frac{1}{(n\ln(n))^x}$$
Now this series converges for $x>1$.
Can we get a closed form of this function for $x>1$?
1
vote
3
answers
177
views
Find the sum $\sum_{n=1}^\infty\frac{\sqrt{n}}{2^{n/2}}$
Find the sum $\sum_{n=1}^\infty\frac{\sqrt{n}}{2^{n/2}}$. Is it possible to find a closed form for this sum? I was trying to upper bound this sum by another series, but could not find one.
Can anyone ...
4
votes
1
answer
180
views
A new sum that equals $-\frac12 \ln (\frac{\pi}{2})$
(See EDIT)
I found this result while working another problem in two slightly different directions and it kind of took me by surprise.
$$
\begin{align}
&\sum_{n=1}^{\infty}\left(\frac{\ln4n}{4n-1}-\...
7
votes
2
answers
268
views
Possible to get a closed form expression, or an upper bound, for $ f(n)=\sum_{m=1}^\infty \bigg(\frac{m+n}{3}\bigg)^{m+n}\bigg(\frac{1}{m}\bigg)^m$?
Is it possible to get a closed form expression, or an upper bound, for the following function $f$ which is given by an infinite summation:
$$
f(n) = \sum_{m=1}^\infty \bigg(\frac{m+n}{3}\bigg)^{m+n} \...
-1
votes
1
answer
108
views
Does $\sum\limits_{k=2}^{\infty}{\frac{|B_{k}|}{k!}(\cos(n)-1)}$ have a closed form?
I am trying to find a closed form expression of the following sum in terms of $n$ (if it exists) where $B_{k}$ is the $k$th Bernoulli number.
$$\sum_{k=2}^{\infty}{\frac{|{B_{k}|}}{k!}(\cos(n)-1)}$$
...
8
votes
1
answer
259
views
Showing that $\sum_{j=0}^{2n-1}{\cos^n(\frac{j\pi}{2n})(2\cos(\frac{2j\pi}n)+1)\cos(\frac{j\pi}2-\frac{2j\pi}n)}$ is never an integer for $n>10$
I want to show that
$$f(n) = \sum_{j=0}^{2n-1}{\cos^n\left( \frac{j \pi}{2n}\right) \left( 2\cos \left( \frac{2 j \pi}{n} \right) + 1\right) \cos \left( \frac{j \pi}{2} - \frac{2 j \pi}{n} \right)}$$
...
3
votes
1
answer
98
views
How can I sum the series $e^{-2}\frac{(3)^n}{n!}\sum_{k=0}^{\infty}\left ( \frac{1}{2}\right )^k\frac{1}{(k-n)!}$
How can I sum the following series?
$$e^{-2}\frac{(3)^n}{n!}\sum_{k=0}^{\infty}\left ( \frac{1}{2}\right )^k\frac{1}{(k-n)!}$$
I think I can make this sum in the form of exponential expansion but ...
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)\...
1
vote
1
answer
66
views
Values for which this sum can be defined in terms of known constants in a closed form
I'm interested in the sum,
$$\sum_{n=1}^\infty\frac{\zeta(2n)\Gamma(2n)}{\Gamma(2k+2n+2)}x^{2n}$$
Otherwise written as
$$\sum_{n=1}^\infty\frac{\zeta(2n)}{(2n)(2n+1)\cdots(2n+2k+1)}x^{2n}$$
I am ...
3
votes
3
answers
3k
views
Sum of a series $\frac {1}{n^2 - m^2}$ m and n odd, $m \ne n$
I was working on a physics problem, where I encountered the following summation problem:
$$ \sum_{m = 1}^\infty \frac{1}{n^2 - m^2}$$ where m doesn't equal n, and both are odd. n is a fixed constant
...
4
votes
1
answer
197
views
Formula for $\sum\limits_{j=1}^{m-1}\frac{1}{\sin^{2p}(\frac{j\pi}{m})}$
Let $m\geq 2$ be an integer, then there is the well known formula
$$\sum\limits_{j=1}^{m-1}\frac{1}{\sin^2(\frac{j\pi}{m})}=\frac{m^2-1}{3},$$
I'm interested in similar equations for the following ...
2
votes
0
answers
92
views
How to find $\sum_{n \in \mathbb Z_+} \frac{2^{n-1}}{2^{2^n}}$?
I'm trying to calculte the measure of a fat Cantor set, but run into this question:
How to find $$\sum_{n \in \mathbb Z_+} \frac{2^{n-1}}{2^{2^n}}$$
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 ...