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Series

Summation operator

The sigma operator $\sum$ expresses the sum of many similar terms. The sign is the capital Greek letter sigma and corresponds to S for sum.

The sum of a collection of numbers $x_1,x_2,\dots,$ between indices $m$ and $n$ is $$\sum_{i=m}^n x_i = x_m + x_{m+1} + x_{m+2} + \dots + x_{n}$$

• The counting begins with the lower bound $m$.
• The counting ends with the upper bound $n$ (included).
• The index of summation $i$ varies between the bounds to select the items to be added.

For $u_n = 2n$, we have $\displaystyle\sum_{i=3}^5 u_n = 6+8+10=24$.

Here are some properties of the $\Sigma$ operator, where $a$ is a constant: \begin{gather*} \sum_{i=1}^n(x_i+y_i)=\sum_{i=1}^n x_i+\sum_{i=1}^n y_i,\quad \sum_{i=1}^n a x_i=a\sum_{i=1}^n x_i\\ \sum_{i=1}^n a=a+a+\cdots+a=na \end{gather*}

The first two properties say that the summation operator is linear.

Partial sum of sequences

Informally, a series is the sum of the first terms of a sequence.

Strictly speaking, an infinite series is the sum of all the terms of the sequence. The partial sum is the sum of all the terms up to a given index. The term series is used either to designate the partial sum or the corresponding infinite series.

The partial sum for a sequence $(u_n)$ is defined by $$S_n = \sum_{k=1}^n u_k = u_1+\dots+ u_{n-1}+ u_n.$$ It is itself a sequence.

If $u_n=1$ for all $n$, we get $S_1 = 1$, $S_2 = 1+1 = 2$, etc. so that $S_n = n$.

The values of the sequence can be calculated from the partial sums via the formula $u_n = S_{n} - S_{n-1}.$

$1/2^n$.
The partial sum is shown using segments of circles." />

Infinite series and convergence

We say that a series is convergent if the sequence of the partial sums is convergent (i.e. has a limit when $n\to\infty$).The limit is denoted by $$\sum_{k=1}^\infty u_k =\lim_{n\to+\infty} \sum_{k=1}^n u_k.$$

The limit of the partial sum is called the infinite series. It is de-fined only if the series is convergent.

The series $\sum 1$ is not convergent, because the sequence of the partial sum $S_n = \sum_{k=1}^n 1 = n$ is not convergent.

The series $\sum 2^{-n}$ is convergent and its value is $1$. Indeed, as we will see, the general term of the (geometric) series is $$S_n = \sum_{k=1}^n 2^{-k} = \frac{2^{-1}-2^{-n}}{1-2^{-1}} = 1-2^{1-n}.$$ Thus the sequence $S_n$ converges to $1$.

$0$.
The partial sum converges to $1$" />

Telescoping series

A series is telescoping if the partial sums have a limited number of terms after cancellation.

For instance, if $u_n = v_{n+1} - v_n$, the partial sum is $$S_n = \sum_{k=1}^{n} u_k = (v_2 - v_1) + (v_3-v_2) +\dots + (v_{n+1}-v_n) = v_{n+1}-v_1.$$ If $v_n$ converges, the series converges to $\displaystyle\sum_{k=1}^{\infty} u_k =\lim_{n\to+\infty} v_n - v_1$.

The series $\sum \frac{1}{n(n+1)}$ is a telescoping series as the general term is $$u_n = \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}.$$ The partial sum is $S_n = \sum_{k=1}^n u_k = 1 - \frac{1}{n+1}$ and it converges to 1.

Method of difference for series

Recall that a telescoping series is one where the partial sums have many terms that cancel.

The series $\sum \frac{1}{n(n+1)}$ is a telescoping series as the general term is $$u_n = \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}.$$ The partial sum is $S_n = \sum_{k=1}^n u_k = 1 - \frac{1}{n+1}$ and it con-verges to 1.

The method of difference consists in writing the general term of a sequence $u_n$ as the difference of two successive terms of another sequence ($1/n$ in the example above). This facilitates the computation of the series $\sum u_n$ as a telescoping series.