Tool to perform formal calculations with the summation operator Σ ∑ (sigma), allowing arithmetic additions from 1 to n.

Summation Σ - dCode

Tag(s) : Arithmetics

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In arithmetic, the summation notation $ \sum_1^n $ (with the Greek letter sigma uppercase) allows to compute a finite addition going from $ 1 $ to $ n $ with an increment of 1 (by default).

__Example:__ The sum of the first $ 5 $ integers $$ 1 + 2 + 3 + 4 + 5 = \sum_{i=1}^{5} i $$

Sometimes the sum can be simplified with a formula:

__Example:__ The sum of the $ n $ first integers $$ 1 + 2 + 3 + \cdots + (n-1) + n = \sum_{i=1}^{n} i = \frac{n(n+1)}{2} $$

The calculation by hand is time-consuming, some sums are interesting to learn/know.

Sometimes the sum does not converge to a value, it can diverge and not have a formula to calculate it.

The notation $ \sum_1^\infty $ (sometimes shortened in $ \sum $) indicates the computation of an infinite addition going from $ 1 $ to infinity $ \infty $ with an increment of 1 (by default).

__Example:__ The sum of the inverses of the $ n $ prime squares (Basel problem) $$ \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots = \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} $$

The demonstration of these sums often involves a limit calculation or a series expansion.

There are many mathematical series (finite or infinite) useful to learn and know, here is a non-exhaustive list:

— Faulhaber's formulas (sum of p-th powers of the first m integers):

$$ \sum_{k=1}^m k = 1 + 2 + \cdots + m = \frac{m(m+1)}{2} $$

$$ \sum_{k=1}^m k^2 = \frac{m(m+1)(2m+1)}{6} = \frac{m^3}{3}+\frac{m^2}{2}+\frac{m}{6} $$

— The particular values of the Riemann Zeta function:

$$ \sum^{\infty}_{k=1} \frac{1}{k^2} = \zeta(2) = 1 + \frac{1}{2^2} + \frac{1}{3^2} + \cdots = \frac{\pi^2}{6} $$

$$ \sum^{\infty}_{k=1} \frac{1}{k^4} = \zeta(4) = 1 + \frac{1}{2^4} + \frac{1}{3^4} + \cdots = \frac{\pi^4}{90} $$

— Powers and exponentials

$$ \sum^{\infty}_{k=0} \frac{1}{k!} = \frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\cdots = e $$

$$ \sum_{k=0}^{n} z^k = \frac{1-z^{n+1}}{1-z} $$

$$ \sum_{k=0}^\infty \frac{z^k}{k!} = e^z $$

$$ \sum_{k=0}^\infty k \frac{z^k}{k!} = z e^z $$

— Trigonometric functions

$$ \sum_{k=0}^\infty \frac{(-1)^k z^{2k+1}}{(2k+1)!} = \sin(z) $$

$$ \sum_{k=0}^\infty \frac{(-1)^k z^{2k}}{(2k)!} = \cos(z) $$

$$ \sum_{k=0}^n {n \choose k} = 2^n $$

$$ \sum_{k=0}^\infty {\alpha \choose k} z^k = (1+z)^\alpha , \quad |z|<1 $$

$$ \sum^{\infty}_{k=1} \frac{(-1)^{k+1}}{k} = \frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots = \ln(2) $$

$$ \sum^{\infty}_{k=1} \frac{(-1)^{k+1}}{2k-1} = \frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\cdots = \frac{\pi}{4} $$

The notation $ \sum \sum $ is read $ \sum \left( \sum \right) $ so the inner sum (inside the parenthesis) is calculated first first, before the outer sum is calculated in a second step.

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Cite as source (bibliography):

*Summation Σ* on dCode.fr [online website], retrieved on 2024-09-13,

sum,summation,sigma,addition,plus,total,serie,nested

https://www.dcode.fr/summation-calculator

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