# Series

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.

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." />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 $$1$$" />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.

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.