Mathematics Stack Exchange
Mathematics Stack Exchange

Questions tagged [summation]

Questions about evaluating summations, especially finite summations. For infinite series, please consider the (sequences-and-series) tag instead.

The notation $\sum\limits_{i=1}^na_i$ means $a_1+\ldots +a_n$.

Use for sums of infinite series and questions of convergence; use for questions about finite sums and simplification of expressions involving sums.

17770 questions
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Pictorial derivation of sum of cubes

In the following picture, the formula for sum of squares of first $n$ natural numbers is derived using a clever construction of 3 triangles. This can be seen as a generalization to the legendary Gauss's trick, or perhaps as an application of group…
Taiben
  • 320
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how to solve $\sum _{m=0}^{k-1}mC_{k-1}^{m}C_{N-k}^{m}$?

solving $$\sum _{m=0}^{k-1} mC_{k-1}^m C_{N-k}^m$$ the solution seems to be $$\frac {\left( N-2\right) !} {\left( k-2\right) !\left( N-k-1\right) !}$$ according to some clue from the other problem. struggle with this the whole afternoon, please…
Jason Hu
  • 349
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How do I separate the summand of a double summation?

For example, if I have: $$\sum_{i=1}^n \sum_{j=1}^n (x_i^2 - 2x_ix_j + x_j^2)$$ How would I separate these terms? Unfortunately, we've dived into proofs using summation without much background information on how it works. My best (but I think…
mrt
  • 145
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determining $n$ in a given sequence $\frac{1+3+5+...+(2n-1)}{2+4+6+...+(2n)} =\frac{2011}{2012} $

Given that: $$\frac{1+3+5+...+(2n-1)}{2+4+6+...+(2n)} =\frac{2011}{2012} $$ Determine $n$. The memorandum says the answer is 2011 but how is that so? Where did I go wrong?
Aspiring Mathlete
  • 2,312
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How to prove that $\sum_{k=0}^n \binom{2n+1}{2k+1} = 4^{n}$

My question is how to prove that $\sum_{k=0}^n \binom{2n+1}{2k+1} = 4^{n}$ . I'm not good at operating on the sign of sum, so please, try to explain me that as clearly as possible. Thanks :)
Jerry
  • 31
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Exact value of $\sum\limits_{n=1}^\infty(-1)^{n(n+1)/2}/n$?

Wolfram is not computing it properly. What is the exact value of $$\sum_{n=1}^\infty\frac{(-1)^{n(n+1)/2}}{n}?$$ How to avoid imaginary $i$ coming from the exponent?
waqar ahmad
  • 109
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How find this sum $\sum_{n=0}^{\infty}(-1)^{n}\frac{n+1}{(2n+1)!}$

Find this follow sum $$\sum_{n=0}^{\infty}(-1)^{n}\dfrac{n+1}{(2n+1)!}$$ My…
user94270
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5 answers

How find this sum $f(n)=\sum_{i=1}^{n}\dfrac{\binom{n}{i}}{i}$

Find this sum closed form $$f(n)=\sum_{i=1}^{n}\dfrac{\binom{n}{i}}{i}$$ My idea: since…
math110
  • 93,304
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how to prove that $\sum_{k=1}^{m}k!k=(m+1)!-1$ without induction?

how to prove that $$\sum_{k=1}^{m}k!k=(m+1)!-1$$ without induction ? my only try is to put $k!=\Gamma(k+1)$ then use geometric series with some steps but I got complicated integral If any one can solve it using my way or similar way using calculus…
user130806
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2 answers

How find this sum $\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)(n+3)\cdots(n+2014)}$

Find the sum $$\sum_{n=1}^{\infty}\dfrac{1}{n(n+1)(n+2)(n+3)\cdots(n+2014)}$$ My idea: since $$\dfrac{1}{n(n+1)(n+2)\cdots(n+2014)}=\dfrac{1}{2014}\left(\dfrac{1}{n(n+1)(n+2)\cdots(n+2013)}-\dfrac{1}{(n+1)(n+2)\cdots(n+2014)}\right)?$$
math110
  • 93,304
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How do I compute this triple summation?

$$\sum_{i=0}^{n-1} \sum_{j=0}^{i-1} \sum_{k=0}^{j-1} i + j + k$$ The question is looking for a $\Theta(g(n))$ function to represent this summation, but I am uncertain how to go about computing triple summations. P.S. This is not homework. It was…
Public Display Name
  • 233
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Is there anything significant about this seemingly arbitrary constant?

$$\sum^N\frac{1}{n^n} = 1 / 1^1 + 1/2^2 + 1/3^3 + 1/4^4 \ + \dotsb= 1.291285997... \text{ for }\ N \to \infty $$ Maybe this is related to the Riemann-Zeta function? I'm taking a wild guess.
Armend Veseli
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Explicit closed form $\sum \frac{1}{an^2+bn+c}$

I know that $\sum_{n\geq 1} \frac{1}{n^2}=\pi^2/6$, is there a simple way to get an explicit closed form $\sum \frac{1}{an^2+bn+c}$, where $a,b,c$ are integers, $a\neq 0$?
Anonymous
  • 65
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Calculate a multiple sum of inverse integers.

The question is to calculate a following sum: \begin{equation} {\mathcal S}_p(n) :=\sum\limits_{1\le j_1 < j_2 < \dots
Przemo
  • 11,331
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commuting sums and products

We place ourselves in the ring of formal power series. Why is it that : $$\sum_{(k_{1},...,k_{d})\in \Bbb{N}^{d}}\prod_{i=1}^dT^{k_{i}m_{i}}=\prod_{i=1}^d\sum_{k\in \Bbb{N}}T^{km_{i}}$$ I guess that developing everything works but is there an easier…
Minuteman
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