6 Infinite Series and Convergence Tests

A structured guide to defining infinite series, recognizing important forms, and selecting convergence tests through clear criteria and worked examples.

Foundations: Partial Sums and Convergence

An infinite series is defined through its sequence of finite partial sums. For

\sum_{n=1}^{\infty}a_n=a_1+a_2+a_3+\cdots,

the partial sum through term $N$ is

S_N=\sum_{n=1}^{N}a_n.

The series converges to a finite number $S$ when

\lim_{N\to\infty}S_N=S.

If the partial sums do not approach a finite limit, the series diverges. A necessary first check is the : if

\lim_{n\to\infty}a_n\ne0,

then $\sum a_n$ diverges. The converse is not true. For example, $\frac{1}{n}\to0$ , but the harmonic series $\sum_{n=1}^{\infty}\frac{1}{n}$ diverges.

Takeaway: Terms approaching zero are necessary for convergence, but this condition alone does not prove convergence.

Recognizing and Evaluating

A has a constant ratio between consecutive terms:

\sum_{n=0}^{\infty}ar^n=a+ar+ar^2+\cdots.

Its finite partial sums are

S_N=a\frac{1-r^{N+1}}{1-r},\qquad r\ne1.

When $|r|<1$ , the factor $r^{N+1}$ approaches zero, so

\sum_{n=0}^{\infty}ar^n=\frac{a}{1-r}.

When $|r|\ge1$ , the diverges. For example,

\sum_{n=1}^{\infty}3\left(\frac{1}{4}\right)^{n-1}

has initial term $3$ and common ratio $\frac{1}{4}$ . Therefore,

\sum_{n=1}^{\infty}3\left(\frac{1}{4}\right)^{n-1}=\frac{3}{1-\frac{1}{4}}=4.

When the index or exponent is shifted, write out the first few terms to identify the initial term and common ratio.

Takeaway: Identify $a$ and $r$ , verify that $|r|<1$ , and then use $\frac{a}{1-r}$ .

Cancellation in

A is evaluated by expanding a partial sum and canceling adjacent terms. For

\sum_{n=1}^{\infty}(b_n-b_{n+1}),

the partial sum through term $N$ is

S_N=(b_1-b_2)+(b_2-b_3)+\cdots+(b_N-b_{N+1})=b_1-b_{N+1}.

If $b_{N+1}\to L$ , then the series sums to $b_1-L$ .

For example, partial fractions give

\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}.

Thus,

S_N=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\cdots+\left(\frac{1}{N}-\frac{1}{N+1}\right)=1-\frac{1}{N+1}.

Taking the limit yields

\sum_{n=1}^{\infty}\frac{1}{n(n+1)}=1.

Takeaway: Rewrite terms so cancellation is explicit, simplify the finite partial sum, and only then take its limit.

Comparison Tests and Known Benchmarks

Comparison methods are most useful for series with nonnegative terms. If $0\le a_n\le b_n$ eventually, convergence of $\sum b_n$ implies convergence of $\sum a_n$ . Conversely, divergence of $\sum a_n$ implies divergence of $\sum b_n$ .

The is useful when two positive terms have the same dominant behavior. If

L=\lim_{n\to\infty}\frac{a_n}{b_n}

satisfies $0<L<\infty$ , then $\sum a_n$ and $\sum b_n$ have the same convergence behavior.

Consider

\sum_{n=1}^{\infty}\frac{3n^2+1}{n^3-2}.

Compare it with the harmonic series using $b_n=\frac{1}{n}$ :

\lim_{n\to\infty}\frac{(3n^2+1)/(n^3-2)}{1/n}=\lim_{n\to\infty}\frac{3n^3+n}{n^3-2}=3.

Because the limit is finite and positive, the given series has the same behavior as $\sum\frac{1}{n}$ , so it diverges.

Takeaway: Choose a benchmark series whose behavior is known and whose terms match the dominant part of the target terms.

The and

The applies when $a_n=f(n)$ , with $f$ positive, continuous, and decreasing for all sufficiently large $x$ . Then

\sum_{n=N}^{\infty}a_n\quad\text{and}\quad\int_N^{\infty}f(x)\,dx

either both converge or both diverge. The integral determines convergence behavior, not usually the exact sum of the series.

The resulting criterion is

\sum_{n=1}^{\infty}\frac{1}{n^p},

which converges when $p>1$ and diverges when $p\le1$ . The case $p=1$ is the divergent harmonic series.

For

\sum_{n=2}^{\infty}\frac{1}{n\ln n},

use $f(x)=\frac{1}{x\ln x}$ . Since

\int_2^{\infty}\frac{dx}{x\ln x}=\lim_{b\to\infty}\left[\ln(\ln x)\right]_2^b=\infty,

the series diverges.

Takeaway: Verify the hypotheses, translate the term into a function, and analyze the corresponding improper integral.

Alternating Series and Remainder Bounds

An alternating series changes signs from term to term, commonly in the form

\sum_{n=1}^{\infty}(-1)^{n-1}b_n=b_1-b_2+b_3-b_4+\cdots,

where $b_n\ge0$ . The guarantees convergence when the magnitudes eventually decrease,

b_{n+1}\le b_n,

and approach zero,

\lim_{n\to\infty}b_n=0.

For the alternating harmonic series,

\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n},

these conditions hold, so the series converges. However, its absolute-value series is $\sum\frac{1}{n}$ , which diverges. Therefore, it is rather than .

The test also gives an error estimate. If $R_N$ is the remainder after $N$ terms, then

|R_N|\le b_{N+1}.

Takeaway: Alternation can produce convergence through cancellation, but test the absolute-value series separately when the type of convergence matters.

Ratio and Root Tests

The examines

L=\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|.

If $L<1$ , the series converges absolutely.
If $L>1$ , including $L=\infty$ , the series diverges.
If $L=1$ , the test is inconclusive.

This test is especially effective for factorials, exponentials, and products of consecutive factors. For

\sum_{n=1}^{\infty}\frac{n!}{3^n},

let $a_n=\frac{n!}{3^n}$ . Then

\left|\frac{a_{n+1}}{a_n}\right|=\frac{n+1}{3}\to\infty,

so the series diverges. Its terms also fail to approach zero.

The examines

L=\lim_{n\to\infty}\sqrt[n]{|a_n|}.

It has the same three outcomes: absolute convergence for $L<1$ , divergence for $L>1$ , and no conclusion for $L=1$ . It is particularly effective when a term is raised to the $n$ th power.

For

\sum_{n=1}^{\infty}\left(\frac{2n+1}{5n}\right)^n,

\sqrt[n]{\left|\left(\frac{2n+1}{5n}\right)^n\right|}=\frac{2n+1}{5n}\to\frac{2}{5}<1,

so the series converges absolutely.

Takeaway: Use the for factorial or product structure, the for nth-power structure, and treat a limiting value of $1$ as inconclusive.

Absolute Versus Conditional Convergence

A series is when

\sum_{n=1}^{\infty}|a_n|

converges. Absolute convergence always implies ordinary convergence. A series is when $\sum a_n$ converges but $\sum|a_n|$ diverges.

For example,

\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}

is : the alternating series converges, but its absolute-value series is the divergent harmonic series.

In contrast,

\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}

is because

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

converges by the criterion.

When signs vary, applying a convergence test to $|a_n|$ is a standard way to determine absolute convergence.

Takeaway: First establish convergence of the original series; then test the absolute-value series to distinguish absolute from conditional convergence.

A Practical Strategy for Choosing a Test

Use the following decision process for an unfamiliar series:

Check the terms. If $a_n\not\to0$ , apply the .
Look for a familiar form, such as a or .
Look for cancellation in partial sums; partial fractions may reveal a .
For positive terms, try direct comparison, the , or the .
For alternating signs, apply the , then test $\sum|a_n|$ if absolute or conditional convergence is requested.
For factorials or products of consecutive factors, try the .
For expressions raised to the $n$ th power, try the .
If a ratio-test or root-test limit equals $1$ , do not draw a conclusion; select a different method.

A complete solution should state the test, verify its relevant hypotheses, compute the decisive limit or comparison, and clearly conclude convergence, divergence, absolute convergence, or conditional convergence.

Final takeaway: Test selection depends on structure: geometric behavior suggests the geometric-series formula, cancellation suggests telescoping, positive dominant behavior suggests comparison or integration, alternating signs suggest the , factorials suggest ratios, and nth powers suggest roots.