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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 ...
mcmat23's user avatar
  • 1,070
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 ...
hellofriends's user avatar
  • 1,940
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 $x$. $$...
Max's user avatar
  • 910
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{(...
Giulio Binosi's user avatar
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}...
keagan_callis's user avatar
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), \...
user268193's user avatar
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}}{\...
Harmonic Sun's user avatar
  • 1,984
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
entechnic's user avatar
  • 496
29 votes
1 answer
570 views

A nice Combinatorial Identity

I am trying to show that $\forall N\in\mathbb{N}$, $$\sum\limits_{n=0}^{N}\sum\limits_{k=0}^{N}\frac{\left(-1\right)^{n+k}}{n+k+1}{N\choose n}{N\choose k}{N+n\choose n}{N+k\choose k}=\frac{1}{2N+1}$$ ...
Harmonic Sun's user avatar
  • 1,984
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}^{...
ABCDEFG user157844's user avatar
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 ...
Nilotpal Sinha's user avatar
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 ...
Y.L's user avatar
  • 187
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) = ...
user avatar
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 ...
user321627's user avatar
  • 2,624
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?
Averroes2's user avatar
  • 1,157

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