Convergence Tests for Series | Sebastián Fernández Rivera

What follows presents several convergence tests for numerical series. See also: Important topics and examples, Series of functions. Let $\{x_{n}\}_{n=0}^\infty$ be a sequence. Consider

\[S_{n} = \sum_{i=1}^{n} x_{k} = x_{1} + x_{2} +\dots+x_{n}\]

We say that $\sum_{k=0}^{\infty} x_{k}$ converges if $\lim_{ n \to \infty } S_{n}$ exists

Cauchy condition

The series $\sum_{n=1}^\infty x_{n}$ converges if and only if for every $\varepsilon > 0$ there exists $N \in \mathbb{N}$ such that

\[\lvert S_{n} - S_{m} \rvert < \varepsilon \quad \text{for all \quad$n,m \geq N$}\] \[\iff \lvert x_{n+1} + x_{n+2} + \dots + x_{m} \rvert < \varepsilon \quad \text{if \quad$m > n$}\]

The series converges if and only if it is Cauchy.

Example

Consider $x_{n} = \frac{1}{n}, n\geq 1$. We will prove that $S_{2n} - S_{n} \geq \frac{1}{2}$. Note that

\[\begin{aligned} S_{2n} - S_{n} &= \left( 1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{2n} \right) -\left( 1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{n} \right) \\ &= \frac{1}{n+1} + \frac{1}{n+2} +\dots +\frac{1}{2n-1} + \frac{1}{2n} \\ &\geq \frac{1}{2n} + \frac{1}{2n} + \dots +\frac{1}{2n} + \frac{1}{2n} = \frac{n}{2n} = \frac{1}{2}. \end{aligned}\]

Hence the Cauchy condition fails for $\varepsilon < \frac{1}{2}$.

Consider the case $x_{n} \geq 0$. Then

\[S_{n+1} = S_{n} + x_{n+1}\geq S_{n}\] \[\implies S_{n+1} - S_{n} = x_{n+1}.\]

Lemma (Boundedness of partial sums):

Let $\{x_{n}\}_{n=1}^\infty$ be such that $x_{n} \geq 0$ . Then

$\sum_{n=0}^{\infty}x_{n}$ converges if there exists $M$ such that $\lvert S_{n} \rvert \leq M$ for every $n$.
Otherwise the $S_{n}$ are unbounded, i.e., $\lim_{ n \to \infty }S_{n} = \infty$.

The p-series test

Example

Consider

\[\sum_{n=1}^{\infty} \frac{1}{n} = +\infty\]

Proof: If $x_{n} \geq 0$, then the $S_{n}$ are increasing. Hence $S_{n}$ converges if and only if it is bounded. Take $M > 0$; there exists $N$ such that $S_{N} > M$. Then, for $n\geq N$,

\[S_{n} \geq S_{N} > M.\]

Example

Consider $x_{n} = \frac{1}{n^2}$, that is,

\[\sum_{n=1}^{\infty} \frac{1}{n^2}.\]

We know that $\lim_{ n \to \infty }S_{n} = \infty$ or $\lim_{ n \to \infty } S_{n} = l \in \mathbb{R}$. Note that:

\[\begin{aligned} S_{1} =& 1 \\ S_{2} =& 1 + \frac{1}{2^2} \\ S_{3} =& 1 + \frac{1}{2^2} + \frac{1}{3^2} \leq 1 + \frac{1}{2^2} + \frac{1}{2^2} = 1+\frac{1}{2} \\ S_{7} =& 1 + \frac{1}{2^2} + \frac{1}{3^2} + \dots + \frac{1}{7^2} \leq 1 + \frac{1}{2} + \frac{1}{4^{2}} + \frac{1}{4^{2}} + \frac{1}{4^{2}} + \frac{1}{4^{2}} \\ =&1 + \frac{1}{2} + \left( \frac{1}{2} \right)^2. \end{aligned}\]

Consider

\[S_{2^j -1} = S_{n_{j}}.\]

We will prove that $S_{n_{j}} \leq \sum_{n=0}^{j-1} \left( \frac{1}{2} \right)^n \quad \text{(*)}$. We have already done the base case. For the inductive step, suppose that $\text{(*)}$ holds. We must show that

\[S_{n_{j+1}} \leq \sum_{n=0}^{j+1} \left( \frac{1}{2} \right)^n.\]

Note that

\[\begin{aligned} S_{n_{j+1}} &= S_{n_{j}} + x_{2^j} + x_{2^{j}+1} + \dots + x_{2^{j+1}-1} \\ &\leq \sum_{n=0}^{j} \left( \frac{1}{2} \right)^n + \frac{2^j}{(2^j)^{2}} = \sum_{n=0}^{j-1} \left( \frac{1}{2} \right)^n + \left( \frac{1}{2} \right)^j = \sum_{n=1}^{j} \left( \frac{1}{2} \right)^n \end{aligned}\]

Since

\[\sum_{n=0}^\infty \left( \frac{1}{2} \right)^n = \frac{1}{1-\frac{1}{2}}.\]

it is bounded, so $S_{2^{j+1} - 1} \leq M$, and therefore it converges.

We know that

\[\sum_{n=1}^\infty \frac{1}{n} = +\infty\] \[\sum_{n=1}^\infty \frac{1}{n^{2}} \quad \text{converges}.\]

Let $0<p<1$. Then

\[n > n^p \implies \frac{1}{n^p} > \frac{1}{n} \implies \sum_{k=0}^\infty \frac{1}{k^p} > \sum_{k=0}^\infty \frac{1}{k} \rightarrow \infty\]

Therefore it diverges to infinity. If $p>2$, then

\[\frac{1}{n^2} > \frac{1}{n^p},\]

so $\sum_{n=1}^\infty \frac{1}{n ^p}$ converges.

In general, if $p > 1$, the series $\sum_{n=1}^\infty \frac{1}{n^p}$ converges.

Exercise:

Let $x_{n} = \frac{1}{n^{p}}$, with $p > 1$. Show that

\[S_{2^j - 1} \leq \sum_{i=0}^{j-1} \frac{1}{2^{j-1}}\]

Theorem (Comparison)

Let $\{x_{n}\}_{n=0}^\infty$ and $\{z_{n}\}_{n=0}^\infty$ be such that $0\leq x_{n}\leq z_{n}$ for every $n\geq 1$. Then

If $\sum_{n=1}^\infty z_{n}$ converges, then $\sum_{n=1}^\infty x_{n}$ converges.
If $\sum_{n=1}^\infty x_{n}$ diverges, then $\sum_{n=1}^\infty z_{n}$ diverges.

Example

Consider

\[x_{n} = \frac{1}{\sqrt{ n^{3}+1 }} \leq \frac{1}{\sqrt{ n^{3} }} = \frac{1}{n^\frac{3}{2}}\]

Therefore $\sum_{n=0}^\infty x_{n}$ converges.

Lemma (Limit comparison)

Let $\{x_{n}\}_{n=1}^\infty$ and $\{z_{n}\}_{n=1}^\infty$ be sequences such that $x_{n} \geq 0$ and $z_{n} \geq 0$ for $n\geq1$.

If $\lim_{ n \to \infty } \frac{x_{n}}{z_{n}} = l \neq 0$, then either both converge or both diverge to infinity.
If $\lim_{ n \to \infty } \frac{x_{n}}{z_{n}} = 0$,
- then $\sum_{n=0}^\infty x_{n}= +\infty$ implies $\sum_{n=0}^\infty z_{n}= +\infty$.
- then, if $\sum_{n=0}^\infty z_{n}$ converges then $\sum_{n=0}^\infty x_{n}$ converges
If $\lim_{ n \to \infty } \frac{x_{n}}{z_{n}} = \infty$,
- then $\sum_{n=0}^\infty z_{n}= +\infty$ implies $\sum_{n=0}^\infty x_{n}= +\infty$.
- then, if $\sum_{n=0}^\infty x_{n}$ converges then $\sum_{n=0}^\infty z_{n}$ converges

Proof of (a): Assume that

\[\lim_{ n \to \infty } \frac{x_{n}}{z_{n}} = l > 0.\]

Given $\varepsilon > 0$, there exists $N \in \mathbb{N}$ such that for every $n \geq N$

\[l - \varepsilon < \frac{x_{n}}{z_{n}} < l + \varepsilon \implies (l-\varepsilon) z_{n} < x_{n} < (l+\varepsilon) z_{n}.\]

Taking $\varepsilon$ such that $0<\varepsilon< \frac{l}{2}$, we get $0 \leq (l-\varepsilon) z_{n} < x_{n}$. Note that

\[0 \leq (l - \varepsilon) \sum_{k=1}^\infty z_{k} \leq \sum_{k=1}^\infty x_{k} \leq (l+\varepsilon) \sum_{n=0}^\infty z_{n}\]

If $\sum_{k=0}^\infty x_{k}$ converges, then by comparison $\sum_{k=0}^\infty z_{k}$

Example

$\sum_{n=0}^\infty \frac{1}{n^{5}-n^{3}+1}$ behaves like $\sum_{n=1}^\infty \frac{1}{n^{5}}$ since

\[\lim_{ n \to \infty } \frac{\left( \frac{1}{n^{5}-n^{3}+1} \right)}{\left( \frac{1}{n^{5}} \right)} = \lim_{ n \to \infty } \frac{n^{5}}{n^{5}-n^{3}+1} = 1\]

Example

$\sum_{n=0}^\infty \frac{2^{3n}+n^{5}}{10^{n}}$ behaves like $\sum_{n=1}^\infty \left( \frac{8}{10} \right)^{n}$ since

\[\lim_{ n \to \infty } \frac{\frac{2^{3n}+n^{5}}{10^{n}}}{\left( \frac{8}{10} \right)^{n}} = 1\]

Lemma (Root test)

Let $\{x_{n}\}_{n=1}^\infty$ be a sequence. Consider

\[\lim_{ n \to \infty } \sqrt[n]{ \lvert x_{n} \rvert }.\]

If $\lim_{ n \to \infty } \sqrt[n]{ \lvert x_{n} \rvert } = l < 1$, then $\sum_{n=0}^\infty \lvert x_{n} \rvert$ converges.

Proof: Let $\varepsilon > 0$ be such that $0 < l+\varepsilon < 1$. Then there exists $N \in \mathbb{N}$ such that for every $n \geq \mathbb{N}$

\[l-\varepsilon < \sqrt[n]{ \lvert x_{n} \rvert } < l + \varepsilon \implies \lvert x_{n} \rvert <(l+\varepsilon)^n\]

for $n \geq N$. Hence the series converges by comparison.

If $\lim_{ n \to \infty } \sqrt[n]{ \lvert x_{n} \rvert } = l > 1$, then $\sum_{n=0}^\infty \lvert x_{n} \rvert +\infty.$

Example

For $a>0$ let $x_{n}=\frac{a^{n}}{n^{n}}$, with $n\geq{1}$. Then

\[\lim_{ n \to \infty } \sqrt[n]{\frac{a^{n}}{n^{n}}} = \frac{a}{n} = 0.\]

Therefore $\sum_{n=1}^\infty \frac{a^{n}}{n^{n}}$ converges.

Lemma (Ratio test)

Let $\{x_{n}\}_{n=1}^\infty$ be a sequence. Consider

\[\lim_{ n \to \infty } \frac{\lvert x_{n+1} \rvert}{\lvert x_{n} \rvert}\]

If $\lim_{ n \to \infty } \frac{\lvert x_{n+1} \rvert}{\lvert x_{n} \rvert} = l < 1$, then $\sum_{n=0}^\infty \lvert x_{n} \rvert$ converges.

If $\lim_{ n \to \infty } \frac{\lvert x_{n+1} \rvert}{\lvert x_{n} \rvert} = l >1$, then $\sum_{n=0}^\infty \lvert x_{n} \rvert = +\infty$ .

Proof: Let $\varepsilon > 0$ be such that $l-\varepsilon > 1$. Then there exists $N \in \mathbb{N}$ such that for every $n \geq \mathbb{N}$

\[l-\varepsilon < \frac{\lvert x_{n+1} \rvert}{\lvert x_{n} \rvert} \implies (l-\varepsilon) \lvert x_{n} \rvert < \lvert x_{n+1} \rvert\]

Then, iterating,

\[\lvert x_{N+k} \rvert \geq (l-\varepsilon)^k \lvert x_{N} \rvert\]

Since $(l-\varepsilon) > 1$, $\sum_{k=1}^\infty(l-\varepsilon)^k = +\infty$. Hence by comparison, $\sum_{n=0}^\infty \lvert x_{n} \rvert = +\infty$.

Example

Consider $x_{n} = \frac{a^n}{n!}$ with $a>0$. Note that

\[\lim_{ n \to \infty } \left\lvert \frac{a_{n+1}}{a_{n}} \right\rvert = \lim_{ n \to \infty } \frac{a}{n+1} = 0.\]

Hence $\sum_{n=0}^\infty x_{n}$ converges.

Theorem (Absolute convergence):

Let $\{x_{n}\}_{n=1}^\infty$ be a sequence. If $\sum_{n=1}^\infty \lvert x_{n} \rvert$ converges, then $\sum_{n=1}^\infty x_{n}$ converges.

Proof: Let $\{x_{n}\}_{n=1}^\infty$ be a sequence. Then

\[\lvert x_{n} \rvert - x_{n} \leq 2 \lvert x_{n} \rvert\]

for every $n \geq 1$. Assume that $\sum_{n=0}^\infty \lvert x_{n} \rvert$ converges to L. Then

\[\sum_{n=0}^\infty 2 \lvert x_{n} \rvert = 2L.\]

We conclude then that $\sum_{n=0}^\infty \lvert x_{n} \rvert - x_{n}$ converges. Let

\[S_{n} = \sum_{k=0}^n \lvert x_{k} \rvert - x_{k},\] \[\hat{S}_{n} = \sum_{k=0}^{n} \lvert x_{k} \rvert\]

Then $\hat{S}_{n} - S_{n} = \sum_{k=0}^{n} x_{k}$. Therefore $\sum_{n=0}^\infty x_{n}$ converges.

Theorem (Convergence of a product of series)

Consider two sequences $\{x_{n}\}_{n=0}^\infty$ and $\{y_{n}\}_{n=0}^\infty$. If

$\lvert y_{k} \rvert \leq M$ for $k \geq 1$.
$\sum_{k=1}^\infty \lvert x_{k} \rvert$ converges, then $\sum_{k=1}^\infty \lvert x_{k} y _{k} \rvert$ converges.

Proof: Consider two sequences $\{x_{n}\}_{n=1}^\infty$ and $\{y_{n}\}_{n=1}^\infty$. We want to study the convergence of $\sum_{n=1}^\infty x_{n} y_{n}$. Assume that $\lvert y_{k} \rvert \leq M$ for $k \geq 1$ and that $\sum_{k=1}^\infty \lvert x_{k} \rvert$ converges. Note that

\[\sum_{k=1}^\infty \lvert x_{k}y_{k} \rvert \leq \sum_{k=1}^\infty M \lvert x_{k} \rvert = M \sum_{k=1}^\infty \lvert x_{k} \rvert\]

Conclude that $\sum_{k=1}^\infty \lvert x_{k}y_{k} \rvert$ converges, and hence $\sum_{k=1}^\infty x_{k} y_{k}$ converges.

Example

$x_{k} = \frac{\sin k}{k^2}$. Note that $\lvert \sin k \rvert \leq 1$ and $\sum_{k=1}^\infty \frac{1}{k^{2}}$ converges

A more general trick: Consider $\sum_{k=1}^{n} x_{k} y_{k}$. Take $S_{k} = \sum_{k=1}^{n} y_{n}$. Then $y_{k} = S_{k} - S_{k-1}$. Expanding:

\[\sum_{k=1}^{n} x_{k} y_{k} = \sum_{k=1}^{n} x_{k} (S_{k} - S_{k-1}) = \sum_{k=1}^{n} x_{k} S_{k} - \sum_{k=0}^{n} x_{k+1} S_{k} = \sum_{k=1}^{n-1} S_{k}(x_{k} - x_{k+1}) + x_{n} S_{n} - x_{1}S_{0}.\]

In fact, for $m \leq n$, we have that

\[\sum_{k=m}^{n} x_{k}y_{k} = \sum_{k=m}^{n} (x_{k}-x_{k+1}) S_{k} + x_{n}S_{n} - x_{m}S_{m-1}.\]

Theorem (Dirichlet’s test)

Let $\{x_{n}\}_{n=1}^\infty$ and $\{y_{n}\}_{n=1}^\infty$ be such that

$x_{n} \geq x_{n+1}$ for every $n \in \mathbb{N}$,
$\lim_{ n \to \infty } x_{n} = 0$,
there exists $M \in \mathbb{R}$ such that for every $n \in \mathbb{N}$, $\lvert \sum_{k=1}^n y_{k} \rvert \leq M$ for every $n \in \mathbb{N}$. Then $\sum_{n=1}^\infty x_{n}y_{n}$ converges.

Proof: Let $k \leq l$, then

\[\begin{aligned} \left\lvert \sum_{n=k}^{l} x_{n}y_{n} \right\rvert &\leq \sum_{n=k}^{l-1} \lvert x_{n}-x_{n+1} \rvert \lvert S_{n} \rvert +\lvert x_{l} \rvert \lvert S_{l} \rvert + \lvert x_{k} \rvert \lvert S_{k-1} \rvert \\ &\leq \sum_{n=k}^{l-1} (x_{n}-x_{n+1}) M + \lvert x_{l} \rvert M + \lvert x_{k} \rvert M \\ &=M(x_{k}-x_{l}) + Mx_{l} + Mx_{k} = 2Mx_{k} \end{aligned}\]

since $x_{n}\geq_{0}$. Moreover, since $\lim_{ n \to \infty } x_{n} = 0$, there exists $N \in \mathbb{N}$ such that $\lvert x_{k} \rvert < \frac{\varepsilon}{2M}$ if $k \geq N$, for a fixed and arbitrary $\varepsilon > 0$. Hence, if $l \geq l \geq N$, we have that

\[\left\lvert \sum_{n=k}^{l} x_{n} y_{n} \right\rvert < \varepsilon.\]

Therefore the series is Cauchy and converges.

Example

Consider $\sum_{n=1}^\infty \frac{(-1)^{n}}{n}$. Take $x_{n} = \frac{1}{n}$, $y_{n} = (-1)^{n}$. Then

\[S_{n} = \sum_{k=1}^{n} y_{k},\quad \text{In particular, }S_{1} =-1, S_{2} = 0 , S_{3} = -1,\dots\]

Hence $S_{n}$ is bounded. Therefore $\sum_{n=1}^\infty \frac{(-1)^{n}}{n}$ converges.

Lemma (Alternating series convergence)

Let $\{x_{n}\}_{n=1}^\infty$ be such that $\lim_{ n \to \infty } x_{n} = 0$ and $x_{n+1} \leq x_{n}$ for every $n\geq1$. Then $\sum_{n=1}^{\infty} (-1)^{n} x_{n}$ converges.

Example

$x_{n} = \frac{1}{\sqrt{ n }}$, $x_{n}=\frac{1}{\ln(n+1)}$.

Consider $y_{n} = \cos(nx)$, with $x \in [0,2\pi)$. We want to bound $\sum_{k=1}^\infty \cos(nx)$ for $x \neq 0$ and $x \neq \pi$. Note that

\[\begin{aligned} \sin\left( \left( k-\frac{1}{2} \right)x \right) - \sin\left( \left( k+\frac{1}{2} \right)x \right) &= 2 \cos(kx) \sin\left( \frac{x}{2} \right) \\ \iff \cos(kx) &= \frac{1}{2\sin\left( \frac{x}{2} \right)}\left[ \sin\left( \left( k+\frac{1}{2} x \right) \right) - \sin\left( \left( k-\frac{1}{2} \right) x \right) \right] \end{aligned}\]

Then,

\[\begin{aligned} \left\lvert \sum_{k=1}^{n} \cos(kx) \right\rvert &= \frac{1}{\left\lvert 2 \sin\left( \frac{x}{2} \right) \right\rvert } \left\lvert \sum_{k=1}^{n} \left[ \sin\left( \left( k+\frac{1}{2} \right) x \right) - \sin\left( \left( k-\frac{1}{2} \right) x \right)\right] \right\rvert \\ &=\frac{1}{\left\lvert 2 \sin\left( \frac{x}{2} \right) \right\rvert } \underbrace{ \left\lvert \sin\left( \left( n+\frac{1}{2} \right)x \right) -\sin\left( \frac{1}{2} x \right) \right\rvert }_{ \leq 2 \text{ since }-1 \leq \sin(x) \leq 1 } \\ &\leq \frac{1}{\lvert \sin\left( \frac{x}{2} \right) \rvert }. \end{aligned}\]

Theorem (Abel’s test)

Let $\{x_{n}\}_{n=1}^\infty$ and $\{y_{n}\}_{n=1}^\infty$ be sequences such that

$\{x_{n}\}_{n=1}^\infty$ is monotone and convergent.
$\sum_{n=1}^\infty y_{n}$ converges. Then $\sum_{n=1}^\infty x_{n} y_{n}$ converges.

Proof: Consider the following cases:

Case 1: The $x_{n}$ are decreasing. Let $L = \lim_{ n \to \infty } x_{n}$. Define $z_{n} = x_{n} - L$. Then $z_{n}$ is decreasing and $\lim_{ n \to \infty }z_{n} = 0$. Therefore $\sum_{n=1}^\infty y_{n} z_{n}$ converges. Note that

\[\sum_{n=1}^{m} y_{n} z_{n} + \underbrace{ \sum_{n=1}^{m} y_{n} L }_{\text{convergent} } = \sum_{n=1}^{m} y_{n} x_{n}.\]

Conclude that $\sum_{n=1}^\infty x_{n} y_{n}$ converges.

Case 2: The $x_{n}$ are increasing. Let $L = \lim_{ n \to \infty } x_{n}$. Define $z_{n} = L -x_{n}$. Then $z_{n}$ is decreasing and $\lim_{ n \to \infty }z_{n} = 0$. Therefore $\sum_{n=1}^\infty y_{n} z_{n}$ converges. Note that

\[- \sum_{n=1}^{m} y_{n} z_{n} + \underbrace{ \sum_{n=1}^{m} y_{n} L }_{\text{convergent} } = \sum_{n=1}^{m} y_{n} x_{n}.\]

Conclude that $\sum_{n=1}^\infty x_{n} y_{n}$ converges.

Example

\[\sum_{n=2}^\infty \frac{1}{n^{2}} \ln\left( 1-\frac{1}{n} \right)\]

Take $y_{n} = \frac{1}{n^{2}}$ and $x_{n} = \ln\left( 1-\frac{1}{n} \right)$. We know that $\sum_{n=2}^\infty \frac{1}{n^{2}}$ converges. Moreover

\[\lim_{ n \to \infty } x_{n} = \ln(1) = 0.\]

Since $1-\frac{1}{n}$ is increasing, $x_{n}$ is increasing. Conclude by Abel’s test that the series converges.

Example

Consider $\sum_{n=1}^\infty (-1)^{n} \frac{\ln(n)}{n}$. Note that $\left( \frac{\ln(n)}{n} \right)' =\frac{1-\ln(x)}{x^{2}} \leq 0$ if $x \geq e$. Then $\frac{\ln(n)}{n}$ is decreasing for $n \geq 3$ (we only need it to be eventually decreasing). Therefore $\sum_{n=2}^\infty (-1)^{n} \frac{\ln(n)}{n}$ converges.

Theorem (Raabe’s test)

Let $\{x_{n}\}_{n=1}^\infty$ be a sequence such that $x_{n}\neq_{0}$ for every $n \in \mathbb{N}$. Consider

\[\lim_{ n \to \infty } n\left( 1 - \frac{\lvert x_{n+1}\rvert}{\lvert x_{n} \rvert } \right) = L.\]

Then

$\sum_{n=1}^{\infty} \lvert x_{n} \rvert$ converges if $L >1$.
$\sum_{n=1}^\infty \lvert x_{n} \rvert$ diverges if $L < 1$.

Proof: Given $\varepsilon > 0$, there exists $N \in \mathbb{N}$ such that

\[L - \varepsilon \leq n\left( 1 - \left\lvert \frac{x_{n+1}}{x_{n}} \right\rvert \right) \leq L +\varepsilon\]

for $n \geq N$. Assume $L > 1$. Take $\varepsilon>0$ such that $L -\varepsilon > 1$. Now

\[\begin{aligned} \frac{L-\varepsilon}{n} &\leq 1 - \left\lvert \frac{x_{n+1}}{x_{n}} \right\rvert \\ \iff \left\lvert \frac{x_{n+1}}{x_{n}} \right\rvert &\leq 1 - \frac{l-\varepsilon}{n} \\ \iff n \left\lvert \frac{x_{n+1}}{x_{n}} \right\rvert &\leq n - (l -\varepsilon) \\ \iff n \lvert x_{n+1} \rvert &\leq n \lvert x_{n} \rvert -(l-\varepsilon) \lvert x_{n} \rvert \\ \iff n \lvert x_{n+1} \rvert &\leq (n-1) \lvert x_{n} \rvert -(L - \varepsilon - 1) \lvert x_{n} \rvert \end{aligned}\]

Then, for $n\geq2$,

\[0 \leq (l-\varepsilon - 1) \lvert x_{n} \rvert \leq (n-1) \lvert x_{n} \rvert - n \lvert x_{n+1} \rvert.\]

Hence, if $S =\sum_{n=1}^\infty (n-1) \lvert x_{n} \rvert - n \lvert x_{n+1} \rvert$ converges then $\sum_{n=1}^\infty\lvert x_{n} \rvert$ converges. Now, $\sum_{n=1}^{k} (n-1) \lvert x_{k} \rvert - n \lvert x_{n+1} \rvert = -k \lvert x_{k+1} \rvert$. Then S converges if $\lim_{ n \to \infty } k \lvert x_{k+1} \rvert$ exists. But $(n-1) \lvert x_{n} \rvert - n \lvert x_{n+1} \rvert$, i.e., the sequence is decreasing and bounded below (since it is positive). Therefore, conclude that $\sum_{n=1}^\infty \lvert x_{n} \rvert$ converges.

Example

Consider $x_{n} = \frac{1\cdot 3 \cdot 5 \cdot \dots (2n+1)}{2 \cdot 4 \cdot 6 \cdot... \cdot (2n+2)}$. Note that

\[\frac{x_{n+1}}{x_{n}} = \frac{\frac{1\cdot 3 \cdot 5 \cdot \dots (2n+3)}{2 \cdot 4 \cdot 6 \cdot... \cdot (2n+2)}}{ \frac{1\cdot 3 \cdot 5 \cdot \dots (2n+4)}{2 \cdot 4 \cdot 6 \cdot... \cdot (2n+2)}} = \frac{2n+3}{2n+4} \underset{n\rightarrow \infty}{\longrightarrow} 1.\]

Applying Raabe’s test:

\[n\left( 1-\frac{x_{n+1}}{x_{n}} \right) = n\left( 1 - \frac{2n+3}{2n+4} \right) = n\left( \frac{1}{2n+4} \right) \underset{n \rightarrow \infty}{\longrightarrow} \frac{1}{2}.\]

Therefore it diverges.

Theorem (Cauchy condensation)

Let $\{a_{n}\}_{n=1}^\infty$ be decreasing and positive. Then $\sum_{n=1}^\infty a_{n}$ converges if and only if $\sum_{n=1}^\infty 2^{n} a_{2^{n}}$ converges

Proof: Let $\{a_{n}\}_{n=1}^\infty$ be a decreasing, positive sequence. Note that

\[\begin{aligned} 2a_{2} &\leq a_{1}+a_{2} &&\leq 2a_{1} \\ 2a_{4} &\leq a_{3}+a_{4} &&\leq 2a_{3} \\ 2^{2} a_{8} &\leq a_{5}+a_{6}+a_{7}+a_{8} &&\leq 2^{2} a_{5} \\ 2^{3} a_{16} &\leq a_{9} + a_{10} + \dots + a_{16} &&\leq 2^{3}a_{9} \\ & \vdots && \\\ 2^{n-1} a_{2^{n}} &\leq a_{2^{n-1}+1} + \dots + a_{2^{n}} &&\leq 2^{n-1} a_{2^{n-1}+1}. \end{aligned}\]

We want $S_{n} = \sum_{k=1}^{n} a_{k}$ to converge when $\{a_{k}\}_{k=1}^\infty$ is decreasing and positive. Since $S_{n}$ is increasing, because $\{a_{k}\}_{k=1}^\infty$ is positive, there are 2 possibilities:

$S_{n}$ converges if it is bounded.
$\lim_{ n \to \infty } S_{n} = +\infty$ if it is not.

Consider $S_{2^{n}} = a_{1} + a_{2} + a_{4} + \dots + a_{2^{n}}$. We know that, being increasing, $S_{n}$ converges if and only if $S_{2^{n}}$ converges. Now, note that

\[\begin{aligned} S_{2^{n}} &= a_{1}+a_{2}+\dots+a_{2^{n}} \\ &= (a_{1}+a_{2}) + (a_{3} + a_{4}) + (a_{3} + \dots+a_{8}) + \dots+(a_{2^{n-1}+1} + \dots + a_{2^{n}}), \end{aligned}\]

so that, by the expansion above, we have that

\[\begin{aligned} 2a_{2} + 2a_{4}+2^{2} a_{8} + \dots +2^{n-1} a_{2^{n}} &= \frac{1}{2} \sum_{k=2}^{n} 2^{k} a_{2^{k}} + 2a_{2} \\ &\leq S_{2^{n}} \\ & \leq 2a_{1} + 2a_{3} + 2^{2} a_{5} + \dots +2^{n-1} a_{2^{n-1}+ 1} \\ &\leq 2a_{1} + 2a_{2} + 2^{2} a_{4} + \dots + 2^{n-1} a_{2^{n-1}} \\ &= 2a_{1} + \sum_{k=1}^{n-1} 2^{k} a_{2^{k}}. \end{aligned}\]

This test is useful for handling logarithms inside series.

Example

Consider $\sum_{n=2}^\infty \frac{1}{n(\ln n)^{2}}$. Take $\sum_{n=2}^\infty 2^{n} \left(\frac{1}{2^{n}} \frac{1}{(\ln2^{n})^{2}} \right) = \sum_{n=1}^\infty \frac{1}{(n\ln 2)^{2}} = \frac{1}{(\ln 2)^{2}} \sum_{n=2}^\infty \frac{1}{n^{2}}$, which converges by the p-series test.

Theorem (Absolute convergence and permutations)

Let $\{a_{n}\}_{n=1}^\infty$ be such that $\sum_{n=0}^\infty \lvert a_{n} \rvert$ converges and let $\phi: \mathbb{N} \to \mathbb{N}$. Then $\sum_{n=0}^\infty a_{\phi(n)}$ converges and moreover $\sum_{n=0}^\infty a_{\phi(n)} = \sum_{n=0}^\infty a_{n}$.

Proof: Define $S_{k} =\sum_{n=0}^{k} a_{n}$ , $\tilde{S_{k}} = \sum_{n=0}^{k} \lvert a_{n} \rvert$ and $u_{k} = \sum_{n=0}^{k}. a_{\phi(n)}$. We know that $\sum_{n=0}^\infty \lvert a_{n} \rvert$ converges, i.e., being Cauchy we have that, given $\varepsilon > 0$, there exists $N \in \mathbb{N}$ such that for all $n,m\geq N$

\[\lvert \tilde{S_{n}} - \tilde{S_{m}} \rvert < \varepsilon.\]

Then $\lvert a_{n+1} \rvert + \lvert a_{n+2} \rvert + \dots + \lvert a_{m} \rvert < \varepsilon$ Let $\ell = \lim_{ n \to \infty } S_{n}$. Now take $N_{1}$ such that $\lvert S_{k} - \ell \rvert < \varepsilon$ TO BE COMPLETED.

Worked example 1

Consider the series $\sum_{n=1}^\infty \frac{1}{\sqrt{ n(n+1) }}$. Note that $\frac{1}{\sqrt{ n(n+1) }} \approx \frac{1}{\sqrt{ n \cdot n }} = \frac{1}{n}$. Since

\[\lim_{ n \to \infty } \frac{\frac{1}{\sqrt{ n(n+1) }}}{\frac{1}{n}} = 1\]

and $\sum_{n=1}^\infty \frac{1}{n}$ diverges, the given series diverges.

Worked example 2

Consider the series $\sum_{n=1}^\infty \sin\left( \frac{1}{n^{p}} \right)$ for $p>1$. Note that $0 \leq \sin(x) \leq x$ for every $x\geq 0$. Since $0 \leq \frac{1}{n^{p}}$, we have that $0\leq \sin\left( \frac{1}{n^{p}} \right) \leq \frac{1}{n^{p}}$. Therefore the series converges by comparison.

Worked example 3

Consider the series $\sum_{n=1}^{\infty} \frac{n^{n}}{(n+1)^{n+1}}$. Note that

\[\frac{n^{n}}{(n+1)^{n+1}} = \frac{1}{n+1} \frac{1}{\left( 1+\frac{1}{n} \right)^{n}} \approx \frac{1}{n+1}\]

Now, $\lim_{ n \to \infty } (1+\frac{1}{n})^{n} = e$. Hence

\[\lim_{ n \to \infty } \frac{\frac{n^{n}}{(n+1)^{n+1}}}{\frac{1}{n+1}} = e\]

so the given series diverges.

Worked example 4

Consider the series $\sum_{n=1}^\infty \frac{\sqrt{ n } - \sqrt{ n+1 }}{\sqrt{ n(n+1) }} = \sum_{n=1}^\infty \frac{1}{\sqrt{ n+1 }} - \frac{1}{\sqrt{ n }}$. This series converges because it is telescoping and $\lim_{ n \to \infty } \frac{1}{\sqrt{ n }} = 0$. In particular, the series converges to $-1$.

Worked example 5

Compute the following limit:

\[\lim_{ n \to \infty } n\left( 1-\left( 1-\frac{1}{2n} \right)^{p} \right).\]