All Questions
56
questions
1
vote
1
answer
129
views
Prove $\sum_{k=j}^{\lfloor n/2\rfloor}\frac1{4^k}\binom{n}{2k}\binom{k}{j}\binom{2k}{k}=\frac1{2^n}\binom{2n-2j}{n-j}\binom{n-j}j$
let $x>1$, $n\in \mathbb N$ and
$$P_{n}(x)=\dfrac{1}{\pi}\int_{0}^{\pi}(x+\sqrt{x^2-1}\cos{t})^ndt.$$
Prove that
$$P_{n}(x)=\dfrac{1}{\pi}\int_{0}^{\pi}\dfrac{1}{\left(x-\sqrt{x^2-1}\cos{t}\right)^{...
- 137
5
votes
2
answers
185
views
Generating Function $\sum_{k=1}^{\infty}\binom{2k}{k}^{-2}x^{k}$
Closed Form For :
$$S=\sum_{k=1}^{\infty}\binom{2k}{k}^{-2}x^{k}$$
Using the Series Expansion for $\arcsin^2(x)$ one can arrive at :
$$\sum_{k=0}^{\infty}\binom{2k}{k}^{-1}x^{k}=\frac{4}{4-x}-4\arcsin\...
- 3,280
8
votes
1
answer
2k
views
Proving $\sum_{n=1}^{\infty}\binom{2n}{n}^2\frac{4H_{2n}-3H_n}{n2^{4n}}=\zeta(2)$
While trying to solve Prove $\int_{0}^{1}\frac1k K(k)\ln\left[\frac{\left(1+k \right)^3}{1-k} \right]\text{d}k=\frac{\pi^3}{4}$
I was able to reduce it to the following form,
$$\sum_{n=1}^{\infty}\...
- 3,280
-2
votes
1
answer
52
views
Conjectured form of $\sum_{r=0}^{\infty}\binom{2r}{r}\frac{r^{n}}{a^{r}}$
I was able to conjecture this result,
$$\sum_{r=0}^{\infty}\binom{2r}{r}\frac{r^{n}}{a^{r}}=(-1)^n\left[\left(x\cdot \frac{d}{dx}\right)^n\left(\frac{1}{\sqrt{1-\frac{4}{x}}}\right)\right]_{x=a}$$
Is ...
- 3,280
0
votes
0
answers
83
views
Re-writing a sum of binomial coefficients as an integral of shifted Legendre polynomials
This is a question regarding the answer presented here.
In order to make this post self-contained, I am wondering if someone can explain why the sum
$$
\sum_{k=0}^{n}(-1)^{n+k}\binom{n}{k}\binom{n+k}{...
user1211323
1
vote
0
answers
47
views
How Can I Compute $\sum_{k=1}^n (-1)^{k-1} \binom{n}{k} \dfrac{1}{k}$? [duplicate]
Proof
$$
\sum_{k=1}^n (-1)^{k-1} \binom{n}{k} \dfrac{1}{k} = 1 + \dfrac{1}{2} + \dots + \dfrac{1}{n}
$$
This might be a silly question asked by someone else but I cannot find it right now.
I tried to ...
- 83
1
vote
2
answers
70
views
Find $\lim\limits_{n\rightarrow\infty}\sqrt[n]{\sum_{k=0}^{n} \frac{(-1)^k}{k+1}\cdot2^{n-k}\cdot\binom{n }{k}}$
Find
$$\lim_{n\rightarrow\infty}\sqrt[n]{\sum\limits_{k=0}^{n} \frac{(-1)^k}{k+1}\cdot2^{n-k}\cdot\binom{n }{k}}$$
My attemt: By the binomia theorem, we have
$$(1-x)^n={\sum_{k=0}^{n} \binom{n }{k} (...
- 3,588
1
vote
1
answer
69
views
Prove that $-\int_0^1 (2(1-2x)^{2n} - 2)dx - \sum_{k=1}^n \int_0^1 {n\choose k} \dfrac{(-4x(1-x))^k}k dx = 2H_{2n} - H_n$
Prove that $-\int_0^1 (2(1-2x)^{2n} - 2)dx - \sum_{k=1}^n \int_0^1 {n\choose k} \dfrac{(-4x(1-x))^k}k dx = 2H_{2n} - H_n$ where $H_n$ is the nth Harmonic number.
For the first part, I think one can ...
- 2,031
1
vote
2
answers
66
views
Closed form for $\sum_{r=0}^{n}(-1)^r {3n+1 \choose 3r+1} \cos ^{3n-3r}(x)\sin^{3r+1} (x)$
I request a closed form for the following sum $$S(n)=\sum_{r=0}^{n}(-1)^r {3n+1 \choose 3r+1} \cos ^{3n-3r}(x)\sin^{3r+1} (x) $$
I tried using De Moivre's theorem $$\cos(nx)+i\sin(nx)=(\cos (x)+i\sin(...
user1103841
6
votes
1
answer
229
views
An alternating sum
I ran into an alternating sum in my research and would like to know if the following identity is true:
$$
\sum_{i = 0}^{\left\lfloor \left(n + 1\right)/2\right\rfloor} \frac{\left(n + 1 - 2i\right)^{n ...
- 147
3
votes
1
answer
75
views
Asymptotic formula of $\sum_{r=1}^{n}\frac{(-1)^{r-1}}{(r-1)!}{n\choose r}$
I need to find the asymptotic equivalence of the sum $$\sum_{r=1}^{n}\frac{(-1)^{r-1}}{(r-1)!}
{n\choose r} $$ where ${n\choose r}$ is the binomial coefficient.
We have the binomial identity $$(1-x)^...
user1094088
3
votes
1
answer
139
views
Simplifying $\sum_{j=0}^{k+1} 2^{k+1-j} \biggl(\frac{(k+1+j)!}{(k+1)!j!}\biggl) = 4^{k+1}$? [closed]
According to Wolfram Alpha,
$$\sum_{j=0}^{k+1} 2^{k+1-j} \biggl(\frac{(k+1+j)!}{(k+1)!j!}\biggl) = 4^{k+1},$$
but I don't see how this sum can be simplified to this. How is it done? Thanks in advance.
- 357
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
4
votes
2
answers
104
views
Prove that $\sum_{k=0}^n(-1)^k{n \choose k}\frac{1}{k+m+1}=\sum_{k=0}^m(-1)^k{m \choose k}\frac{1}{k+n+1}$.
I actually found this question in a calculus exercise, so I thought maybe it is an idea to convert an infinite sum to a Riemannian Integral.
But then, I realized that it was missing the $\lim_{n \...
6
votes
2
answers
296
views
Find a closed-form solution to the following summation
I am solving a summation that appears in a paper, it claims that
$$\sum_{n=1}^{\infty} \binom{2n}{n+k}z^n=\bigg(\frac{4z}{(1+\sqrt{1-4z})^2}\bigg)^k$$
I found this identity here in equation (66)
$$\...
- 515