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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$\sum_{\text{k}\ge0}\frac{\left(-1\right)^\text{k}\left(2\text{k}\right)!}{2^{2\text{k}}\left(\text{k}!\right)^2}=\frac{1}{\sqrt{2}}$?

How can you compute the following summation: $$\sum_{\text{k}\ge0}\frac{\left(-1\right)^\text{k}\left(2\text{k}\right)!}{2^{2\text{k}}\left(\text{k}!\right)^2}=\frac{1}{\sqrt{2}}$$
rtomas
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What $i-1$ means in a summation formula?

What $i-1$ means when comes on the top of $∑$? $$ \sum_{i=1}^{i-1}v_i $$ (Encountered here) If v is: $2, 4, 6$, then the sum will be: $2+4+6-2$? Thanks.
tamara azevedo
  • 11
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Proof of $\sum_{n=0}^{\infty }a_nt^n\sum_{k=0}^{\infty}b_kt^k=\sum_{n=0}^{\infty }(\sum_{k=0}^{\infty}a_{n-k}b_k)t^n$

How to prove this identity? $$\sum_{n=0}^{\infty }a_nt^n\sum_{k=0}^{\infty}b_kt^k=\sum_{n=0}^{\infty }(\sum_{k=0}^{\infty}a_{n-k}b_k)t^n$$ I found it on a textbook but there was no proof...
Costas
  • 123
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How does this summation work?

For $i = 1,2,\ldots,19$ we have: $$\left(\sum_{j=1, j\neq i}^{19}\alpha_j\right)\cdot(1.5) + (\alpha_i)\cdot(3) = 1.5 \implies \left(\sum_{j=1}^{19}\alpha_j\right)(1.5) + (\alpha_i)(1.5) = 1.5$$ Does anyone know how to prove this? Any intuition…
Mahamad A. Kanouté
  • 710
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Is there a short formula for this?

You know this how you can express a sum of consecutive terms like this: $1 + 2 + 3 + ... + n = \frac{n(n + 1)}{2}$? In other words, only using $n$? I would like to know if there is a way to shorten, or to express such a sum of sums: For example, for…
davaisk
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What is the double sum of:

$\sum\limits_{n=0}^\infty \sum\limits_{m=0}^\infty \frac{ \sin[ka(m-n)]}{(m-n)} , m \neq n $ where $k$ and $a$ are constants. How to treat this double sum?
Relax Environment
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What is the sum of:

$\Sigma_{n=0}^{\infty}\Sigma_{m=0}^{\infty} \frac{1}{(m-n)}$ How to treat this double sum?
Relax Environment
  • 17
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How can this sum be rewritten like this?

How can this sum be rewritten like this $1 + 1 + 2 + 2 + · · · + l + l + (l + 1) + (l + 1) + l + ε$ $= 2\sum\limits_{i=1}^{l+1}i+l+ε$ $= (l+1)(l+2)+l+ε $ First of all I dont even understand how we can go from the first step to the second. To make…
JangoCG
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Double Summation that has Same Upper Limit as Index: $\sum\limits_{i=1}^j\sum\limits_{j=1}^5 3ij$

This is the first time I encountered a double summation. $$\sum_{i=1}^j\sum_{j=1}^5 3ij$$ I have solved the right summation but I don't know what to do with the 'j' upper limit of the left summation. Perhaps its order need to be changed first?
JamesLp
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Summation evaluation

If n takes values less than $-1$ how do I evaluate the following summation: $$y[n]= \sum_{k=n+1}^{\infty} a^{n-k}$$ Do I need to switch signs of n because n takes only negative values? I'm sorry if this question seems stupid....I'm not that well…
Orpheus
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Symmetry in summation

Can someone please explain the motivation behind this equality? $\sum_{i=1}^{n} (\sum_{j=1}^{i} + \sum_{j=i}^{n})\frac{1}{ij} = 2\sum_{i=1}^{n} \sum_{j=1}^{i} \frac{1}{ij}$ They explain that it is symmetry of some kind.
yescactus
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Find finite sum $ \sum_{k = 1}^{n} {k2^k}$

How do you find the finite sum of $ \sum_{k = 1}^{n} {k2^k}$ . Any hint? No integration and differentiation.
math123abc
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Signed (Riemann) sums

Thinking about Riemann's rearrangement theorem, I asked myself the following question: for any real number $\alpha >1$ fixed, are there two integers $1 \leqslant M < N$ and a sequence $(\varepsilon_n)_n \in \{-1,0,1\}^{\mathbb{N}}$ such that $$…
Kermatoni
  • 117
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Derive that $\sum_{k=0}^{n-1} k \ 2^{n-(k+1)} = 2^n-n-1$.

$\sum_{k=0}^{n-1} k \ 2^{n-(k+1)} = 2^n-n-1$. How does one show a sum like this?
Labbsserts
  • 377
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Discrete change of multiple variables?

Suppose I have a sum of the form: $$ \sum_{j=0}^n \sum_{i=0}^m c_{i,j}$$ And I sub : $$ j+ i \to x$$ And, $$ i-j \to y$$ Then: $$ \frac{i+j}{2} = x$$ And, $$ \frac{i-j}{2} = y$$ If I sub: $$ \sum_{j=0}^n \sum_{i=0}^m c_{ \frac{x+y}{2} ,…
tryst with freedom
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