All Questions
19
questions
1
vote
1
answer
29
views
Subset of index that minimizes a sum of real values
Given a series of real numbers $c_1, \ldots, c_n$ with $ n \in \mathbb{N} $, is there an algorithm or method to find the subset of indices such that the absolute value of the sum of the values within ...
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
0
answers
97
views
Fractional part of a sum
Define for $n\in\mathbb{N}$ $$S_n=\sum_{r=0}^{n}\binom{n}{r}^2\left(\sum_{k=1}^{n+r}\frac{1}{k^5}\right)$$
I need to find $\{S_n\}$ for $n$ large where $\{x\}$ denotes the fractional part of $x$.
$$...
5
votes
1
answer
285
views
Proving that an alternative sum of factorials is zero
I have to prove that for every $k$ the following differential equation is fulfilled by monomials of even degree less than $2k$:
$$
\sum_{j=1}^ka_j^{(k)}x^{j-1}f^{(j)}(x)=0,
$$
with $a_j^{(k)}:=\frac{(...
1
vote
0
answers
36
views
How to bound $\sum_{ j=1}^n j^{n-j}?$ [duplicate]
I came across this sum in some research I'm doing. I need to bound
$$S_n=\sum_{j=1}^{n}j^{n-j}.$$
One bound I've managed is obtained by doing the following: Divide by n! and observe
$S_n/n!=\sum_{j=1}...
0
votes
1
answer
54
views
Computation of a finite sum
I am working with Gamma functions and I needto compute the following sum:
\begin{align}
\sum_{k=1}^n k\frac{\Gamma(n-k+\alpha)}{\Gamma(n-k+1)}=\sum_{k=1}^nk(n-k+\alpha)(n-k+\alpha-1)\cdots(n-k+1),
\...
2
votes
1
answer
110
views
An impressive combinatorial identity
I'm trying to show that, $\forall n\in\mathbb{N},\forall p\in\mathbb{N^*},$ $$\sum_{k=1}^{n+1}\left(\frac{4^k}{2k-1}{2k\choose k}\frac{\displaystyle{n+k-1\choose n-k+1}{2(n+p-k)\choose n+p-k}}{\...
3
votes
3
answers
210
views
I cannot prove that $ \sum_{k=0}^n \sum_{i=k}^n {n \choose k} {n+1 \choose i+1} = 2^{2n} $
I 've tried to calculate the internal sum first with no success.
$$
\sum_{k=0}^n \Bigg( {n \choose k} \sum_{i=k}^n {n+1 \choose i+1} \Bigg) =
2^{2n}
$$
Thank you in advance
29
votes
1
answer
570
views
A nice Combinatorial Identity
I am trying to show that $\forall N\in\mathbb{N}$,
$$\sum\limits_{n=0}^{N}\sum\limits_{k=0}^{N}\frac{\left(-1\right)^{n+k}}{n+k+1}{N\choose n}{N\choose k}{N+n\choose n}{N+k\choose k}=\frac{1}{2N+1}$$
...
0
votes
1
answer
63
views
When$ f(x) = \sum_{n=0}^{200}a_{n}x^{n}$ satisfy with $f(x)+f(x-1) = (x+1)^{200}$ for all $x\in \mathbb{R}$ , find the value of [detail]
Given a polynomial $ f(x) = \sum_{n=0}^{200}a_{n}x^{n}$ satisfied with equation $f(x)+f(x-1) = (x+1)^{200}$ is true for all $x\in \mathbb{R}$
Question : What is the value of $1+ \frac{2 \sum_{n=51}^{...
3
votes
3
answers
200
views
A conjecture on the sum of binomial coefficients
I am looking for a proof, disproof or counter example of the following claim.
Let $C = \{k_1, k_2, \ldots, \}$ be a strictly increasing infinite sequence of positive integers which have a certain ...
0
votes
1
answer
86
views
Simplify the sum $\sum_{i_1+i_2+...+i_m=z}{x_{1}^{i_1}x_{2}^{i_2}\cdot...\cdot x_{m}^{i_m}}$
I was wondering is it possible to simplify the following sum:
$$\sum_{i_1+i_2+...+i_m=z}{x_{1}^{i_1}x_{2}^{i_2}\cdot...\cdot x_{m}^{i_m}}$$
where $0<x<1$ for all $x$.
Is it possible to lose ...
2
votes
3
answers
396
views
$\sum_{j=1}^n \sum_{k=j+1}^n (1)$
This is an elementary question but somehow I am having a hard time seeing it. Can someone post a step by step why is it that:
$$\sum_{j=1}^n \sum_{k>j} \ (1) = \sum_{j=1}^n \sum_{k=j+1}^n (1) = ...
0
votes
2
answers
156
views
How to show that $\sum_{N=n}^{100} \frac{\frac{4}{9}^N}{(N-n)!(100-N)!} = \frac{\frac{4}{13}^n\frac{13}{9}^{100}}{(100-n)!}$?
I am trying to show that for $n\geq 0$,
$$
\sum_{N=n}^{100} \frac{\left(\frac{4}{9}\right)^N}{(N-n)!(100-N)!} = \frac{\left(\frac{4}{13}\right)^n\left(\frac{13}{9}\right)^{100}}{(100-n)!}
$$
Is ...
3
votes
1
answer
816
views
General formula for this sum $\sum_{k=1}^n\frac{1}{k(k+1)...(k+m)}$
Is there a general formula for
$\sum_{k=1}^n\frac{1}{k(k+1)...(k+m)}$?
I know that the limit is $\frac{1}{mm!}$ but is there a combinatorial expression for this?