The main purpose of this article is to provide examples of convergent series. We also speak of divergent series. Before that, we first give a concise summary of the properties of the series and recall the convergence criteria of the series with the proofs.

In fact, to fully understand the contents of this page, information on convergence sequences is necessary.

## Convergent series, definition, and properties

In the sequel, we recall some generalities about convergent series.

In mathematics, a numerical ** series** is the production of a sequence of real numbers $(u_n)$ and a sequence of partial sums $S_n:=u_0+\cdots+u_n$ for all $n\in\mathbb{N}$.

The series is convergent if the sequence $(S_n)_n$ has a finite limit. It is said to be divergent if the limit of $(S_n)_n$ is $\pm\infty$ or does not exist.

In general, the series associated with a sequence $(u_n)_n$ is denoted by $\sum_{n=0}^{+\infty} u_n,$ or $\sum_{n\ge 0} u_n$ if the sequence $u_n$ is defined for and $n\ge 0$. It is denoted by $\sum_{n=N}^{+\infty} u_n,$ or $\sum_{n\ge N} u_n$ is the sequence $u_n$ is define for any $n\ge N$, for some $N\in\mathbb{N}$. We mention that the convergence of series does not depend on the initial index $N$.

**Examples:** Here we give the classic series that you absolutely must know:

Let $a\in (-1,1)$. Then $\sum_{n=0}^{+\infty}a^n$ is convergent series and \begin{align*} \sum_{n=0}^{+\infty}a^n=\frac{1}{1-a}.\end{align*} In fact, according to the Remarkable identities we have \begin{align*} S_n=1+a+a^2+\cdots+a^n=\frac{1-a^{n+1}}{1-a}.\end{align*} On the other hand, as $a\in (-1,1),$ then the geometric sequence $(a^n)_n$ is convergent and we have $a^{n+1}\to 0$ as $n\to+\infty$. Thus the result follows.*The geometric series:*The series $\sum_{n=1}^{+\infty}\frac{1}{n}$ is divergent and we have \begin{align*}\sum_{n=1}^{+\infty}\frac{1}{n}=+\infty.\end{align*} In fact, by contradiction, assume that this series is convergent. Then there exists a real number $\ell$ such that $S_n\to \ell$ and $n\to +\infty$. Remark that \begin{align*} S_{2n}-S_n=\sum_{k=n+1}^{2n} \frac{1}{k}\ge \sum_{k=n+1}^{2n} \frac{1}{2n}=\frac{1}{2}.\end{align*} But we have also $S_{2n}\to \ell,$ so that by letting $n\to\infty$ in both sides of the above inequality, we obtain $0\ge \frac{1}{2}$. This is a contradiction. Thus the harmonic series is not convergent.*The harmonic series:*

A series $\sum_{n=0}^{+\infty}u_n$ is called** absolutely convergent** if the series $\sum_{n=0}^{+\infty}|u_n|$ is convergent in the above sense.

Remark: Every absolutely convergent series is a convergent series. In fact, denote by $(S_n)_n$ and $(S^|u|_n)_n$ the partial sums of sequences $(u_n)_n$ and $(|u_n|)_n$. By assumption, the sequence is convergent, so it is a Cauchy sequence. Let us know to prove that $(S_n)_n$ is convergent. In fact, for any $p,q\in\mathbb{N}$ with $p>q$, we have $$ |S^u_p-S^u_q|\le \sum^{p}_{k=q+1} |u_k|=S^{|u|}_p-S^{|v|}_q.$$ This implies that $(S^u_n)_n$ is a Cauchy sequence. Thus it converges. This ends the proof.

## Series of positive terms

Let a series $\sum_{n=0}^{n}u_n$ be such that the terms $u_n\ge 0$ for any $n$. As $S_{n+1}-S_n=u_n\ge 0,$ the partial sums sequence $(S_n)_n$ is increasing. Thus the series $\sum_{n=0}^{n}u_n$ is convergent if and only if there exists a real number $M>0$ such that $0\le S_n\le M$ for any $n$.

What you should take into account for a series of positive terms is that the convergence of the series is equivalent to just finding an upper bound of the sequence $(S_n)_n$. We then use this remark to prove the following comparison criteria for convergence and divergence of series.

**Theorem:** Let $\alpha\in (1,+\infty)$. The the series $$ \sum_{n=1}^\infty \frac{1}{n^\alpha}$$ is convergent.

**Proof:** We will use background from improper integrals. We now that for $\alpha>0,$ we have \begin{align*} \int^{+\infty}_1 \frac{1}{x^\alpha}dx=\left[ \frac{x^{1-\alpha}}{1-\alpha}\right]^{x=+\infty}_{x=1}=\frac{1}{\alpha-1}.\end{align*} As the terms of the series $u_n=\frac{1}{n^\alpha}$ are positive, we will use the above integral to fine an upper bounded of the partial sums sequence $$ S_n=1+\frac{1}{2^\alpha}+\cdots+\frac{1}{n^\alpha},\quad \forall n\in\mathbb{N}^\ast.$$ We have \begin{align*} S_n=1+\frac{1}{2^\alpha}+\cdots+\frac{1}{n^\alpha}&<1+\int^n_1 \frac{1}{x^\alpha}dx\cr & < 1+\int^{+\infty}_1 \frac{1}{x^\alpha}dx=1+\frac{1}{\alpha-1}=\frac{\alpha}{\alpha-1}.\end{align*}This ends the proof.

### Comparison test for series

**Proposition:** Convergence of series using comparison test: in fact, assume that we have two sequences $(u_n)_n$ and $(v_n)_n$ such that $0\le u_n\le v_n$ for all $n\ge 0$, then

- if the $\sum_{n=0}^{+\infty}v_n$ converges then the series $\sum_{n=0}^{+\infty}u_n$ also converges.
- The series $\sum_{n=0}^{+\infty}u_n$ diverges implies that the series $\sum_{n=0}^{+\infty}v_n$ diverges as well.

**Proof:** 1- Let denote by $(S^u_n)_n$ and $(S^v_n)_n$ the partial sums sequences associated with the sequence $(u_n)$ and $(v_n)$. As, by assumption, $(S^v_n)_n$ is convergent, then there exists a real $M>0$ such that $S_n^v\le M$. But $S_n^u\le S^v_n\le M$ for any $n$. This implies that the series $\sum_{n=0}^{+\infty}v_n$ is convergent.

2- This is just the contraposition of the first result.

**Examples:** 1- The series $\sum_{n=0}^{+\infty}\sin\left(\frac{1}{2^n}\right)$ is convergent. In fact, using the fact that $|\sin(x)|\le |x|$ for any $x\in \mathbb{R},$ we deduce that $\left|\sin\left(\frac{1}{2^n}\right)\right|\le \left( \frac{1}{2}\right)^n$ for any $n$. We know that the geometric series $\sum_{n=0}^\infty \left( \frac{1}{2}\right)^n$ is convergent. Thus the series $\sum_{n=0}^{+\infty}\sin\left(\frac{1}{2^n}\right)$ is absolutely convergent, then convergent.

2- The series $\sum_{n=1}^{+\infty} \frac{1}{\sqrt{n}}$ is divergent. In fact, for any $n\ge 1,$ we have $\frac{1}{n}\le \frac{1}{\sqrt{n}}$ and the harmonic series $\sum_{n=1}^{+\infty}\frac{1}{n}$ is divergent. The result follows using the above proposition.

3- The series $\sum_{n\ge 0}\frac{n}{n^3+1}$ is convergent. In fact, for any $n\ge 1,$ we have $0<\frac{n}{n^3+1}\le \frac{n}{n^3}=\frac{1}{n^2}$. Thus the convergence follows by the convergence of the series $\sum_{n=1}^\infty \frac{1}{n^2}$. Here we take $\alpha=2$ in the above theorem.

### The equivalence test for convergent series

**Theorem:** Let $(u_n)_n$ and $(u_n)_n$ be a positive equivalent sequences, $u_n\sim v_n$. That is $\frac{u_n}{v_n}\to 0$ as $n\to\infty$. Then the series $\sum_{n=0}^\infty u_n$ and $\sum_{n=0}^\infty v_n$ are of the same nature, both convergent, or both divergent.

**Proof:** By using the sequences limit definition, pour any $\varepsilon\in (0,1)$, there exists $N\in\mathbb{N}$ such that for any $n\ge N$, we have $|\frac{u_n}{v_n}-1|\le \varepsilon$. We also have \begin{align*} (1-\varepsilon) v_n\le u_n\le (1+\varepsilon),\quad \forall n\ge N.\end{align*} The result follows by using the comparaison test, see the previous subsectio.

**Examples:** 1- The series $\sum_{n=0}^{+\infty}\sin(\frac{1}{n})$ is divergent. In fact, we have $\sin(1/n)\ge 0$ because for any $n\ge 1,$ $\frac{1}{n}\in(0,1]\subset [0,\frac{\pi}{2}]$. On the other hand, \begin{align*} \lim_{n\to\infty}\frac{\sin(\frac{1}{n})}{\frac{1}{n}}=1.\end{align*} Thus $\sin(1/n)\sim (1/n)$. The result now follows from the divergence of the harmonic series $\sum_{n\ge 1} \frac{1}{n}$.