Testing for Convergence or Divergence of a Series

TL;DR

• nth-term test for divergence: $\lim_{n\to\infty} a_n \neq 0 \Rightarrow \text{series }\sum a_n \text{ diverges}$
• Convergence/divergence of geometric series: Geometric series $\sum ar^{n-1}$
  • converges if $|r| < 1$
  • diverges if $|r| \geq 1$
• Convergence/divergence of p-series: p-series $\sum \cfrac{1}{n^p}$
  • converges if $p>1$
  • diverges if $p\leq 1$
• Comparison Test: If $0 \leq a_n \leq b_n$, then
  • $\sum b_n < \infty \ \Rightarrow \ \sum a_n < \infty$
  • $\sum a_n = \infty \ \Rightarrow \ \sum b_n = \infty$
• Limit Comparison Test: If $\lim_{n\to\infty} \frac{a_n}{b_n} = c \text{ (}c\text{ is a finite positive number)}$, then both series $\sum a_n$ and $\sum b_n$ either both converge or both diverge
• For a series of positive terms $\sum a_n$ and a positive number $\epsilon < 1$
  • If $\sqrt[n]{a_n}< 1-\epsilon$ for all $n$, then the series $\sum a_n$ converges
  • If $\sqrt[n]{a_n}> 1+\epsilon$ for all $n$, then the series $\sum a_n$ diverges
• Root Test: For a series of positive terms $\sum a_n$, if the limit $\lim_{n\to\infty} \sqrt[n]{a_n} =: r$ exists,
  • the series $\sum a_n$ converges if $r<1$
  • the series $\sum a_n$ diverges if $r>1$
• Ratio Test: For a sequence of positive terms $(a_n)$ and $0 < r < 1$
  • If $a_{n+1}/a_n \leq r$ for all $n$, then the series $\sum a_n$ converges
  • If $a_{n+1}/a_n \geq 1$ for all $n$, then the series $\sum a_n$ diverges
• For a sequence of positive terms $(a_n)$, if the limit $\rho := \lim_{n\to\infty} \cfrac{a_{n+1}}{a_n}$ exists,
  • the series $\sum a_n$ converges if $\rho < 1$
  • the series $\sum a_n$ diverges if $\rho > 1$
• Integral Test: For a continuous, decreasing function $f: [1,\infty) \rightarrow \mathbb{R}$ with $f(x)>0$ always, the series $\sum f(n)$ converges if and only if the integral $\int_1^\infty f(x)\ dx := \lim_{b\to\infty} \int_1^b f(x)\ dx$ converges
• Alternating Series Test: An alternating series $\sum a_n$ converges if the following conditions are met:
  1. The signs of $a_n$ and $a_{n+1}$ are different for all $n$
  2. $|a_n| \geq |a_{n+1}|$ for all $n$
  3. $\lim_{n\to\infty} a_n = 0$
• A series that converges absolutely also converges. The converse is not true.

Prerequisites

• Sequences and Series

Introduction

In the previous post on Sequences and Series, we learned about the definitions of convergence and divergence of series. In this post, we will summarize various methods that can be used to determine the convergence or divergence of series. Generally, determining the convergence or divergence of a series is much easier than calculating the exact sum of the series.

nth-term test for divergence

For a series $\sum a_n$, $a_n$ is called the general term of the series.

The following theorem allows us to easily identify some series that clearly diverge, and therefore, it is wise to check this first when determining the convergence/divergence of a series to avoid wasting time.

nth-term test for divergence
If a series $\sum a_n$ converges, then

\[\lim_{n\to\infty} a_n=0\]

In other words,

\[\lim_{n\to\infty} a_n \neq 0 \Rightarrow \text{series }\sum a_n \text{ diverges}\]

Proof

Let $l$ be the sum of a converging series $\sum a_n$, and let

\[s_n := a_1 + a_2 + \cdots + a_n\]

be the sum of the first $n$ terms. Then,

\[\forall \epsilon > 0,\, \exists N \in \mathbb{N}\ (n > N \Rightarrow |s_n - l| < \epsilon).\]

Therefore, for sufficiently large $n$ (> N),

\[|a_n| = |s_n - s_{n-1}| = |(s_n - l) - (s_{n-1} - l)| \leq |s_n - l| + |s_{n-1} - l| \leq \epsilon + \epsilon = 2\epsilon\]

From the definition of convergence of a sequence,

\[\lim_{n\to\infty} |a_n| = 0. \quad \blacksquare\]

Caution

The converse of this theorem is generally not true. A representative example that demonstrates this is the harmonic series.

The harmonic series is a series obtained from a sequence where each term is the reciprocal of an arithmetic sequence, i.e., a harmonic sequence. A typical harmonic series is

\[H_n := 1 + \frac{1}{2} + \cdots + \frac{1}{n} \quad (n=1,2,3,\dots)\]

This series can be shown to diverge as follows:

\[\begin{align*} \lim_{n\to\infty} H_n &= 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \cdots + \frac{1}{16} + \cdots \\ &> 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{16} + \cdots + \frac{1}{16} + \cdots \\ &= 1 + \frac{1}{2} \qquad\, + \frac{1}{2} \qquad\qquad\qquad\ \ + \frac{1}{2} \qquad\qquad\quad + \frac{1}{2} + \cdots \\ &= \infty. \end{align*}\]

As we can see, despite the divergence of the series $H_n$, the general term $1/n$ converges to 0.

While $\lim_{n\to\infty} a_n \neq 0$ necessarily implies that the series $\sum a_n$ diverges, it is dangerous to assume that the series $\sum a_n$ will converge just because $\lim_{n\to\infty} a_n = 0$. In such cases, other methods should be used to determine convergence/divergence.

Geometric Series

The geometric series obtained from a geometric sequence with first term 1 and common ratio $r$

\[1 + r + r^2 + r^3 + \cdots \label{eqn:geometric_series}\tag{5}\]

is the most important and fundamental series. From the equation

\[(1-r)(1+r+\cdots + r^{n-1}) = 1 - r^n\]

we get

\[1 + r + \cdots + r^{n-1} = \frac{1-r^n}{1-r} = \frac{1}{1-r} - \frac{r^n}{1-r} \qquad (r \neq 1) \label{eqn:sum_of_geometric_series}\tag{6}\]

On the other hand,

\[\lim_{n\to\infty} r^n = 0 \quad \Leftrightarrow \quad |r| < 1\]

Therefore, we know that the necessary and sufficient condition for the geometric series ($\ref{eqn:geometric_series}$) to converge is $|r| < 1$.

Convergence/divergence of geometric series
The geometric series $\sum ar^{n-1}$

• converges if $|r| < 1$
• diverges if $|r| \geq 1$

From this, we obtain

\[1 + r + r^2 + r^3 + \cdots = \frac{1}{1-r} \qquad (|r| < 1) \label{eqn:sum_of_inf_geometric_series}\tag{7}\]

Geometric Series and Approximation

The identity ($\ref{eqn:sum_of_geometric_series}$) is useful for finding an approximation of $\cfrac{1}{1-r}$ when $|r| < 1$.

Substituting $r=-\epsilon$ and $n=2$ into this equation, we get

\[\frac{1}{1+\epsilon} - (1 - \epsilon) = \frac{\epsilon^2}{1 + \epsilon}\]

Therefore, if $0 < \epsilon < 1$,

\[0 < \frac{1}{1 + \epsilon} - (1 - \epsilon) < \epsilon^2\]

so we obtain

\[\frac{1}{1 + \epsilon} \approx (1 - \epsilon) \pm \epsilon^2 \qquad (0 < \epsilon < 1)\]

From this, we can see that for a sufficiently small positive $\epsilon$, $\cfrac{1}{1 + \epsilon}$ can be approximated by $1 - \epsilon$.

p-Series Test

For a positive real number $p$, a series of the following form is called a p-series:

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

Convergence/divergence of p-series
The p-series $\sum \cfrac{1}{n^p}$

• converges if $p>1$
• diverges if $p\leq 1$

In p-series, when $p=1$, it becomes the harmonic series, which we have shown to diverge.
The problem of finding the value of the p-series when $p=2$, i.e., $\sum \cfrac{1}{n^2}$, is called the ‘Basel problem’, named after the hometown of the Bernoulli family, which first proved that this series converges and produced several famous mathematicians over several generations. The answer to this problem is known to be $\cfrac{\pi^2}{6}$.

More generally, the p-series with $p>1$ is called the zeta function. This is one of the special functions introduced by Leonhard Euler in 1740 and later named by Riemann, defined as

\[\zeta(s) := \sum_{n=1}^{\infty} \frac{1}{n^s} \qquad (s>1)\]

This topic deviates somewhat from the main subject of this post, and to be honest, I’m an engineering student, not a mathematician, so I don’t know much about it myself. Therefore, I won’t discuss it here. However, Leonhard Euler showed that the zeta function can also be expressed as an infinite product of prime numbers, known as the Euler Product, and subsequently, the zeta function occupies a central position in several fields under analytic number theory. The Riemann zeta function, which extends the domain of the zeta function to complex numbers, and the important unsolved problem related to it, the Riemann hypothesis, are among them.

Returning to the original topic, the proof of the p-series test requires the Comparison Test and Integral Test, which will be discussed later. However, the convergence/divergence of p-series can be usefully applied in the Comparison Test along with geometric series, which is why it has been intentionally placed near the beginning.

Proof

i) When $p>1$

The integral

\[\int_1^\infty \frac{1}{x^p}\ dx = \left[\frac{1}{-p+1}\frac{1}{x^{p-1}} \right]^\infty_1 = \frac{1}{p-1}\]

converges, so by the Integral Test, we know that the series $\sum \cfrac{1}{n^p}$ also converges.

ii) When $p\leq 1$

In this case,

\[0 \leq \frac{1}{n} \leq \frac{1}{n^p}\]

We know that the harmonic series $\sum \cfrac{1}{n}$ diverges, so by the Comparison Test, we can see that $\sum \cfrac{1}{n^p}$ also diverges.

Conclusion

By i) and ii), the p-series $\sum \cfrac{1}{n^p}$ converges if $p>1$ and diverges if $p \leq 1$. $\blacksquare$

Comparison Test

Jakob Bernoulli’s Comparison Test is useful for determining the convergence/divergence of series of positive terms, which are series whose general terms are non-negative real numbers.

A series of positive terms $\sum a_n$ is an increasing sequence, so unless it diverges to infinity ($\sum a_n = \infty$), it must converge. Therefore, in a series of positive terms, the expression

\[\sum a_n < \infty\]

means that it converges.

Comparison Test
If $0 \leq a_n \leq b_n$, then

• $\sum b_n < \infty \ \Rightarrow \ \sum a_n < \infty$
• $\sum a_n = \infty \ \Rightarrow \ \sum b_n = \infty$

In particular, for series of positive terms that have forms similar to the geometric series $\sum ar^{n-1}$ or p-series $\sum \cfrac{1}{n^p}$ that we looked at earlier, such as $\sum \cfrac{1}{n^2 + n}$, $\sum \cfrac{\log n}{n^3}$, $\sum \cfrac{1}{2^n + 3^n}$, $\sum \cfrac{1}{\sqrt{n}}$, $\sum \sin{\cfrac{1}{n}}$, it’s a good idea to actively try the Comparison Test.

All the other convergence/divergence tests that will be discussed later can be derived from this Comparison Test, and in that sense, the Comparison Test can be considered the most important.

Limit Comparison Test

For series of positive terms $\sum a_n$ and $\sum b_n$, suppose that the dominant terms in the numerator and denominator of the ratio of the general terms $a_n/b_n$ cancel out, resulting in $\lim_{n\to\infty} \cfrac{a_n}{b_n}=c \text{ (}c\text{ is a finite positive number)}$. If we know whether the series $\sum b_n$ converges or diverges, we can use the following Limit Comparison Test.

Limit Comparison Test
If

\[\lim_{n\to\infty} \frac{a_n}{b_n} = c \text{ (}c\text{ is a finite positive number)}\]

then both series $\sum a_n$ and $\sum b_n$ either both converge or both diverge. In other words, $ \sum a_n < \infty \ \Leftrightarrow \ \sum b_n < \infty$.

Root Test

Theorem
For a series of positive terms $\sum a_n$ and a positive number $\epsilon < 1$

• If $\sqrt[n]{a_n}< 1-\epsilon$ for all $n$, then the series $\sum a_n$ converges
• If $\sqrt[n]{a_n}> 1+\epsilon$ for all $n$, then the series $\sum a_n$ diverges

Corollary: Root Test
For a series of positive terms $\sum a_n$, suppose the limit

\[\lim_{n\to\infty} \sqrt[n]{a_n} =: r\]

exists. Then,

• If $r<1$, the series $\sum a_n$ converges
• If $r>1$, the series $\sum a_n$ diverges

In the above corollary, if $r=1$, convergence/divergence cannot be determined, so another method must be used.

Ratio Test

Ratio Test
For a sequence of positive terms $(a_n)$ and $0 < r < 1$

• If $a_{n+1}/a_n \leq r$ for all $n$, then the series $\sum a_n$ converges
• If $a_{n+1}/a_n \geq 1$ for all $n$, then the series $\sum a_n$ diverges

Corollary
For a sequence of positive terms $(a_n)$, suppose the limit $\rho := \lim_{n\to\infty} \cfrac{a_{n+1}}{a_n}$ exists. Then,

• If $\rho < 1$, the series $\sum a_n$ converges
• If $\rho > 1$, the series $\sum a_n$ diverges

Integral Test

The integral method can be used to determine the convergence/divergence of series composed of decreasing positive sequences.

Integral Test
For a continuous, decreasing function $f: [1,\infty) \rightarrow \mathbb{R}$ with $f(x)>0$ always, the series $\sum f(n)$ converges if and only if the integral

\[\int_1^\infty f(x)\ dx := \lim_{b\to\infty} \int_1^b f(x)\ dx\]

converges.

Proof

If the function $f(x)$ is continuous, decreasing, and always positive, the inequality

\[f(n+1) \leq \int_n^{n+1} f(x)\ dx \leq f(n)\]

holds. Adding this inequality from $n=1$ to the general term, we get the inequality

\[f(2) + \cdots + f(n+1) \leq \int_1^{n+1} f(x)\ dx \leq f(1) + \cdots + f(n)\]

Now, using the Comparison Test, we obtain the desired result. $\blacksquare$

Alternating Series

A series $\sum a_n$ where each term $a_n$ is non-zero and has a different sign from the next term $a_{n+1}$, i.e., where positive and negative terms alternate, is called an alternating series.

For alternating series, the following theorem discovered by the German mathematician Gottfried Wilhelm Leibniz can be usefully applied to determine convergence/divergence.

Alternating Series Test
If

1. The signs of $a_n$ and $a_{n+1}$ are different for all $n$,
2. $|a_n| \geq |a_{n+1}|$ for all $n$, and
3. $\lim_{n\to\infty} a_n = 0$,

then the alternating series $\sum a_n$ converges.

Absolute Convergence

For a series $\sum a_n$, if the series $\sum |a_n|$ converges, we say that “the series $\sum a_n$ converges absolutely”.

In this case, the following theorem holds:

Theorem
A series that converges absolutely also converges.

The converse of this theorem does not hold.
When a series converges but does not converge absolutely, we say it “converges conditionally”.

Proof

For a real number $a$, let

\[\begin{align*} a^+ &:= \max\{a,0\} = \frac{1}{2}(|a| + a), \\ a^- &:= -\min\{a,0\} = \frac{1}{2}(|a| - a) \end{align*}\]

Then we have,

\[a = a^+ - a^-, \qquad |a| = a^+ + a^-\]

Since $0 \leq a^\pm \leq |a|$, by the Comparison Test, if the series $\sum |a_n|$ converges, then the series $\sum a_n^+$ and $\sum a_n^-$ also converge. Therefore, by the basic properties of convergent series,

\[\sum a_n = \sum (a_n^+ - a_n^-) = \sum a_n^+ - \sum a_n^-\]

also converges. $\blacksquare$