Numerical series.
Series with non-negative terms.
A series $\sum\limits_{n=1}^{\infty}a_n$ with non-negative terms $(a_n\geq 0, n\in \mathbb{N})$ converges if and only if the sequence of its partial sums is bounded above, that is, there exists a number $M>0$ such that for each $n\in\mathbb{N}$ the inequality $\sum\limits_{n=1}^{\infty}a_n\leq M$ holds.
Comparison tests.
1. If there exists a number $n_0$ such that for all $n\geq n_0$ the inequalities $0\leq a_n\leq b_n$ hold, then convergence of the series $\sum\limits_{n=1}^{\infty}b_n$ implies convergence of the series $\sum\limits_{n=1}^{\infty}a_n$, and divergence of the series $\sum\limits_{n=1}^{\infty}a_n$ implies divergence of the series $\sum\limits_{n=1}^{\infty}b_n$.
2. If $a_n\geq 0,,, b_n> 0$ for all $n\geq n_0$ and there exists a finite non-zero limit $\lim\limits_{n\rightarrow\infty}\frac{a_n}{b_n},$ then the series $\sum\limits_{n=1}^{\infty}a_n$ and $\sum\limits_{n=1}^{\infty}b_n$ either both converge or both diverge.
Example 1.
Using the comparison test, investigate the convergence of the series $\sum\limits_{n=1}^{\infty}\frac{5+3(-1)^{n+1}}{2^n}.$
Solution.
For odd $n$, $a_n=\frac{8}{2^n}$, and for even $n$, $a_n=\frac{2}{2^n}$. Thus, $0 < \frac{1}{2^{n-1}} < a_n < \frac{1}{2^{n-3}}.$
Notice that $\left\{ \frac{1}{2^{n-3}} \right\}$ is a geometric progression with $b_1= 4$ and $q=\frac{1}{2}$. The sum of a geometric progression is calculated using the formula $S=\frac{b_1}{1-q}$. In our case, $S=\frac{4}{1/2}=8$. Therefore, the series $\sum\limits_{n=1}^{\infty}\frac{1}{2^{n-3}}$ converges. By the first convergence test, it follows that $\sum\limits_{n=1}^{\infty}\frac{5+3(-1)^{n+1}}{2^n}$ also converges.
Example 2.
Using the comparison test, investigate the convergence of $\sum\limits_{n=1}^{\infty}\sin\frac{3+(-1)^{n}}{n^2}.$
Solution.
For odd $n$, $a_n=\sin\frac{2}{n^2}$, and for even $n$, $a_n=\sin\frac{4}{n^2}$. Thus, $0<\sin\frac{2}{n^2}<\sin\frac{3+(-1)^{n}}{n^2}<\sin\frac{4}{n^2}.$
Notice that $\lim\limits_{n\rightarrow}\frac{\sin\frac{4}{n^2}}{\frac{4}{n^2}} = 1.$ This means that the series $\sum\limits_{n=1}^{\infty}\sin\frac{4}{n^2}$ and $\sum\limits_{n=1}^{\infty}\frac{4}{n^2}$ converge or diverge simultaneously. Let's consider the series $\frac{4}{n^2}:$
$$\frac{4}{n^2}\leq\frac{4}{n(n-1)}=-\frac{4}{n}+\frac{4}{n-1}=4\left(-\frac{1}{n}+\frac{1}{n-1}\right).$$
Thus, $\sum\limits_{n=1}^{\infty}\frac{4}{n^2}=4\left(-\frac{1}{2}+1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-...\right)$ $S_n=4\left(1-\frac{1}{n}\right),\quad S=\lim\limits_{n\rightarrow\infty} S_n=4<\infty$. Therefore, the series converges. Hence, by the convergence test, it follows that $\sum\limits_{n=1}^{\infty}\sin\frac{3+(-1)^{n}}{n^2}$ also converges.
Integral convergence test for series.
If the function $f(x)$ is non-negative and decreasing on the interval $[a,+\infty),$ where $a\geq 1,$ then the series $\sum\limits_{n=1}^{\infty}f(n)$ and the integral $\int\limits_{a}^{+\infty}f(x)dx$ converge or diverge simultaneously.
Example.
Investigate the convergence of the series $\sum\limits_{n=1}^{\infty}\frac{1}{(1+n)\sqrt{n}}.$
Solution.
We will use the integral convergence test. Consider the integral $\int\limits_1^{+\infty}\frac{1}{(1+x)\sqrt{x}}:$
$\int\limits_1^{+\infty}\frac{1}{(1+x)\sqrt {x}}=[t=\sqrt{x};,,, dx=2tdt]=\lim\limits_{A\rightarrow +\infty}\int\limits_1^A\frac{2tdt}{t(1+t^2)}=\lim\limits_{A\rightarrow +\infty}2 \arctan t|_1^A=$ $=2\cdot\frac{\pi}{2}-2\cdot\frac{\pi}{4}=\frac{\pi}{2}.$
Hence, the integral $\int\limits_1^{+\infty}\frac{1}{(1+x)\sqrt{x}}$ converges. By the integral convergence test for series, it follows that the series $\sum\limits_{n=1}^{\infty} \frac{1}{(1+n)\sqrt{n}}$ also converges.
Cauchy and D'Alembert criteria.
D'Alembert criterion.
If for the series $\sum\limits_{n=1}^{\infty}a_n,$ $a_n>0$ $(n\in\mathbb{N}),$ there exists a limit $\lim\limits_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}= \lambda$, then if $\lambda<1$ the series $\sum\limits_{n=1}^{\infty}a_n$ converges, if $\lambda>1$ it diverges. When $\lambda=1$, the series may converge or diverge.
Cauchy criterion.
If for the series $\sum\limits_{n=1}^{\infty}a_n,$ $a_n\geq 0$ $(n\in\mathbb{N}),$ there exists a limit $\lim\limits_{n\rightarrow\infty}\sqrt[n]{a_n}=\lambda$, then if $\lambda<1$ the series converges, if $\lambda>1$ it diverges.
Examples.
1. Investigate the convergence of the series $\sum\limits_{n=1}^{\infty}a_n$ using the D'Alembert criterion.
$a_n=\frac{n^{10}}{(n+1)!}$
Solution.
$a_{n+1}=\frac{(n+1)^{10}}{(n+1)!(n+2)}$
$$\lim\limits_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}=\lim\limits_{n\rightarrow\infty}\frac{(n+1)^{10}(n+1)!}{(n+1)!(n+2)n^{10}}=\lim\limits_{n\rightarrow\infty}\frac{(n+1)^{10}}{n^{10}(n+2)}=$$ $$\lim\limits_{n\rightarrow\infty}\frac{n^{10}(1+1/n)^{10}}{n^{10}(n+2)}=\lim\limits_{n\rightarrow\infty}\frac{(1+1/n)^{10}}{n+2}=0<1.$$
Therefore, the series converges.
2. Investigate the convergence of the series $\sum\limits_{n=1}^{\infty}a_n$ using the Cauchy criterion.
$a_n=3^{n+1}\left(\frac{n+2}{n+3}\right)^{n^2}$
Solution.
$$\lim\limits_{n\rightarrow\infty}\sqrt[n]{a_n}=\lim\limits_{n\rightarrow\infty}3\sqrt[n]{3}\left(\frac{n+2}{n+3}\right)^n=\lim\limits_{n\rightarrow\infty}3\sqrt[n]{3}\left(\frac{n+3-1}{n+3}\right)^n=$$ $$\lim\limits_{n\rightarrow\infty}3\sqrt[n]{3}\left(\left(1-\frac{1}{n+3}\right)^{-(n+3)+3}\right)^{-1}=\frac{3}{e}\lim\limits_{n\rightarrow\infty}\left(1+\frac{1}{n+3}\right)^{-3}=\frac{3}{e}.$$
Raabe and Gauss criteria.
Raabe criterion.
If $a_n>0 ,, (n\in \mathbb{N})$ and there exists a limit $\lim\limits_{n\rightarrow\infty}n\left(\frac{a_n}{a_{n+1}}-1\right)=q$, then if $q>1$ the series $\sum\limits_{n=1}^{\infty}a_n$ converges and if $q<1$ it diverges.
Gauss criterion.
If $a_n>0,,, (n\in \mathbb{N})$ and $\frac{a_n}{a_{n+1}}=\alpha+\frac{\beta}{n}+\frac{\gamma_n}{n^{1+\delta}},$ where $|\gamma_n|< C< \delta>0, $ then:
a) if $\alpha>1$ the series $\sum\limits_{n=1}^{\infty}a_n$ converges, and if $\alpha<1$ it diverges.
b) if $\alpha=1$ the series converges if $\beta>1$ and diverges if $\beta\leq 1$.
Examples.
1. Investigate the convergence of the series $\sum\limits_{n=1}^{\infty}a_n$ using the Raabe or Gauss criterion.
$a_n=\frac{(2n-1)!!}{(2n)!!}$
Solution.
We will use the Raabe criterion.
$$a_{n+1}=\frac{(2n+1)!!}{(2n+2)!!}=\frac{(2n-1)!!(2n+1)}{(2n)!!(2n+2)}=a_n\frac{2n+1}{2n+2}.$$
$$\lim\limits_{n\rightarrow\infty} n\left(n(\frac{a_n}{a_{n+1}-1})\right)= \lim\limits_{n\rightarrow\infty}n\left(\frac{2n+2}{2n+1}-1\right)=\lim\limits_{n\rightarrow\infty}n\left(\frac{2n+2-2n-1}{2n+1}\right)=$$ $$\lim\limits_{n\rightarrow\infty}\frac{n}{2n+1}=\frac{1}{2}<1.$$
Therefore, the series diverges.
2. Investigate the convergence of the series $\sum\limits_{n=1}^{\infty}a_n$ using the Raabe or Gauss criterion.
$a_n=\left(\frac{(a+1)(a+2)...(a+n)}{(b+1)(b+2)...(b+n)}\right)^{\alpha}$
Solution.
We will apply the Gauss criterion.
$$a_{n+1}=\left(\frac{(a+1)(a+2)...(a+n+1)}{(b+1)(b+2)...(b+n+1)}\right)^{\alpha}=a_n\left(\frac{(a+n+1}{b+n+1}\right)^{\alpha}.$$
$$\frac{a_n}{a_{n+1}}=\left(\frac{b+n+1}{a+n+1}\right)^{\alpha}=\left(1+\frac{b-a}{a+n+1}\right)^{\alpha}=1+\alpha\frac{b-a}{a+n+1}+$$ $$\alpha(\alpha-1)\frac{1}{2}\left(\frac{b-a}{a+n+1}\right)^2+...$$
Therefore, the series converges if $\alpha(b-1)>1$ and diverges if $\alpha(b-a)\leq 1.$
Alternating series.
A series $\sum\limits_{n=1}^{\infty}(-1)^{n-1}a_n= a_1-a_2+a_3+...,$ where $a_n\geq 0$ or $a_n\leq 0$ for all $n\in \mathbb{N}$ is called an alternating series.
Leibniz criterion.
If $\lim\limits_{n\rightarrow\infty}a_n=0$ and for each $n\in \mathbb{N}$ it holds that $a_n\geq a_{n+1}>0$ , then the series $\sum\limits_{n=1}^{\infty}(-1)^{n-1}a_n$ converges. Moreover, $|S-S_n|\leq a_{n+1},$ where $S$ and $S_n$ are the sum and $n$-th partial sum of the series $\sum\limits_{n=1}^{\infty}(-1)^{n-1}a_n.$
Examples.
1. Investigate the convergence of the series $\sum\limits_{n=1}^{\infty}\frac{(-1)^n}{\sqrt[3]{n+1}}.$
Solution.
Let's use the Leibniz criterion. Since the sequence $\left\{ \frac{1}{\sqrt[3]{n+1}} \right\}$ decreases and tends to 0 as $n\rightarrow\infty,$ by the Leibniz criterion, the series converges.
Homework.
Investigate the convergence of the series $\sum\limits_{n=1}^{\infty}a_n$.
1)$a_n=\frac{1}{n^{\alpha}}$ for all values of the parameter $\alpha.$
2) $a_n=\frac{n^3}{3^n}.$
3) $a_n=\frac{1\cdot5\cdot8\cdot...\cdot(3n-1)}{1\cdot6\cdot11\cdot...\cdot(5n-4)}.$
4) $a_n=\left(\frac{3}{n}\right)^n.$
5) $a_n=\left(\frac{n^2+5}{n^2+6}\right)^{n^3}.$
6) $a_n=\left(\frac{(2n+1)!!}{(2n+2)!!}\right)^{\alpha}\frac{1}{n^{\beta}}$ for all values of $\alpha$ and $\beta.$$
7) $\sum\limits_{n=1}^{\infty}\frac{(-1)^{n+1}\ln n}{\sqrt{n}}.$
Tags:
