summation

Summation

Summation notation is simply defining a short-hand notation for adding up the terms in a sequence, for example \[\sum_{n=m}^{p}a_{n}=a_{m}+a_{m+1}+\dotso+a_{p}\].
The index variable is considered as a dummy variable in the sense that it can be changed to any letter without affecting the value of the summation. For example, \[\sum_{n=3}^{6}(2n-1)=\sum_{k=3}^{6}(2k-1)\].

Properties

  • \[\sum_{n=m}^{p}(a_{n}\pm b_{n})=\sum_{n=m}^{p}a_{n}\pm\sum_{n=m}^{p}b_{n}\]
  • \[\sum_{n=m}^{p}c\cdot a_{n}=c\cdot\sum_{n=m}^{p}a_{n}\]
  • \[\sum_{n=m}^{p}a_{n}=\sum_{n=m}^{j}a_{n}+\sum_{n=j+1}^{p}a_{n}\]
  • \[\sum_{n=m}^{p}a_{n}=\sum_{n=m+r}^{p+r}a_{n-r}\]

Summation formulae for \[\Sigma r\], \[\Sigma r^{2}\] and \[\Sigma r^{3}\]

  • \[\sum_{r=1}^{n}r=\frac{n(n+1)}{2}\]
  • \[\sum_{r=1}^{n}r^{2}=\frac{1}{6}n(n+1)(2n+1)\]
  • \[\sum_{r=1}^{n}r^{3}=\frac{1}{4}n^{2}(n+1)^{2}\]

Proof of \[\sum_{r=1}^{n}r^{2}=\frac{1}{6}n(n+1)(2n+1)\] with mathematical induction

\begin{align*} \text{Let } n=1:\\ \sum_{r=1}^{1}r^{2}&=\frac{1(1+1)(2\cdot 1+1)}{6}\\ &=1\\ \text{Let }n=k:\\ \sum_{r=1}^{k}r^{2}&=\frac{k(k+1)(2k+1)}{6}\\ \sum_{r=1}^{k+1}r^{2}&=\sum_{r=1}^{k}r^{2}+(k+1)^{2}\\ &=\frac{k(k+1)(2k+1)+6(k+1)^{2}}{6}\\ &=\frac{(k+1)(k(2k+1)+6(k+1))}{6}\\ &=\frac{(k+1)(2k^{2}+7k+6)}{6}\\ &=\frac{(k+1)(k+2)(2k+3)}{6}\\ &=\frac{(k+1)(k+2)(2(k+1)+1)}{6}\\ \end{align*}

Convergent series

A convergent series is series that tends to a limit, that means that one \[a_{n}\] after the other in the order given by the indices, one gets partial sums that become closer and closer to a given number.
The series \[\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\cdots=\ln(2)\] is a converging series.

Method of differences to find \[n\]

Let's take \[\sum_{r=1}^{n}\frac{1}{r(r+2)}\] as an example. Using partial fraction decomposition,

\begin{align*} \sum_{r=1}^{n}\frac{1}{r(r+2)}&=\frac{1}{2}\cdot \sum_{r=1}^{n}\left( \frac{1}{r}-\frac{1}{r+2} \right)\\ &=\frac{1}{2}\cdot\sum_{r=1}^{n}\left( \left( \frac{1}{1}-\frac{1}{3} \right)+\left( \frac{1}{2}-\frac{1}{4} \right)+\left( \frac{1}{3}-\frac{1}{5} \right)+\left( \frac{1}{4}-\frac{1}{6} \right)+\cdots+\left( \frac{1}{n-2}-\frac{1}{n} \right)+\left( \frac{1}{n-1}-\frac{1}{n+1} \right)+\left( \frac{1}{n}-\frac{1}{n+2} \right) \right)\\ &=\frac{1}{2}\cdot \left( 1+\frac{1}{2}-\frac{1}{n+1}-\frac{1}{n+2} \right)\\ \end{align*}

If \[n\to\infty\], \[\lim_{n\to\infty^{+}}\left( \frac{3}{4}-\frac{1}{2(n+1)}-\frac{1}{2(n+2)} \right)=\frac{3}{4}\].

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