Sequences and Infinite Series

A structured guide to sequences, infinite series, convergence, divergence, and the major tests used to classify series and evaluate their sums.

Sequences and Their Limits

A is an ordered list of numbers, such as $a_1,a_2,a_3,\ldots$ , where $a_n$ is the $n$ th term. A can be specified in several ways.

An explicit formula gives a term directly, for example $a_n=\frac{2n+1}{n+3}$ .
A recursive definition gives initial value or values and a rule for generating later terms, for example $a_1=3$ and $a_{n+1}=2a_n-1$ .
A can also be represented graphically or by a pattern.

To study long-term behavior, examine the limit $\lim_{n\to\infty}a_n$ . If the limit is a finite number $L$ , the converges to $L$ . If no finite limit exists, the diverges. Divergence can involve unbounded growth, unbounded decrease, or oscillation, as in $(-1)^n$ .

For rational expressions involving powers of $n$ , divide the numerator and denominator by the highest power of $n$ . Thus, $\lim_{n\to\infty}\frac{4n^2-n+1}{2n^2+3}=2$ , because the highest-degree terms determine the limit. Exponential growth eventually dominates polynomial growth; for example, an expression involving $2^n$ eventually outgrows one involving $n^5$ .

Limits can be combined when the component limits exist and the resulting operation is defined. In particular, sums and products pass to the limit, and quotients do so when the denominator limit is nonzero.

A useful structural result is the Monotone Convergence Theorem: an increasing bounded above converges, and a decreasing bounded below converges.

Takeaway: First identify how a is defined, then determine whether its terms approach a finite value, diverge without bound, or oscillate.

From Sequences to Series

An adds the terms of a :

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

Because infinitely many terms cannot be added one at a time to completion, define the series through its :

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

The series converges to $S$ exactly when

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

If the fail to approach a finite number, the series diverges. This distinction is essential: a concerns the individual terms, while a series concerns the accumulated .

The provides the first check. If $\lim_{n\to\infty}a_n$ is nonzero or does not exist, then $\sum a_n$ diverges. However, $a_n\to0$ is only necessary, not sufficient. The harmonic series $\sum_{n=1}^{\infty}\frac1n$ diverges even though its terms approach zero.

Changing, removing, or adding finitely many terms may change the numerical sum, but it does not change whether the series converges or diverges.

Takeaway: Always check the term limit first, but never conclude convergence from $a_n\to0$ alone.

Geometric and Telescoping Structure

Some series can be classified or summed by recognizing their structure before using a general test.

A has a constant ratio between consecutive terms:

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

Its finite partial sum is

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

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

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

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

\sum_{n=0}^{\infty}5\left(\frac23\right)^n=\frac{5}{1-2/3}=15.

A has cancellation hidden in its terms. For example,

\frac{1}{n(n+1)}=\frac1n-\frac1{n+1}.

Therefore, its $N$ th partial sum is

S_N=\left(1-\frac12\right)+\left(\frac12-\frac13\right)+\cdots+\left(\frac1N-\frac1{N+1}\right)=1-\frac1{N+1}.

Taking the limit gives

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

When working with telescoping expressions, expand several so that the canceled terms and the surviving endpoint terms are clear.

Takeaway: Look for a constant ratio or cancellation pattern before applying a more general convergence test.

Positive-Term Series and Comparison

For positive-term series, compare the expression with a benchmark whose behavior is known. The p-series

\sum_{n=1}^{\infty}\frac1{n^p}

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

The uses inequalities. If $0\le a_n\le b_n$ eventually and $\sum b_n$ converges, then $\sum a_n$ converges. Conversely, if $0\le a_n\le b_n$ and the smaller series $\sum a_n$ diverges, then the larger series $\sum b_n$ diverges. For example,

0<\frac{1}{n^2+4}\le\frac1{n^2},

so $\sum\frac{1}{n^2+4}$ converges.

The is useful when two positive terms have the same dominant growth. Compute

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

If $0<L<\infty$ , the two series have the same convergence classification. For

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

compare with $1/n$ :

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

Because the limit is positive and finite, the series behaves like the divergent harmonic series and therefore diverges.

The applies when $a_n=f(n)$ , where $f$ is positive, continuous, and decreasing eventually. The series and the improper integral $\int_N^{\infty}f(x)\,dx$ then have the same convergence behavior. Applying it to $f(x)=1/x^p$ reproduces the p-series classification. Applying it to $f(x)=1/(x\ln x)$ gives

\int_2^{\infty}\frac{dx}{x\ln x}=\infty,

so $\sum_{n=2}^{\infty}\frac{1}{n\ln n}$ diverges.

Takeaway: Match the form of a positive-term series to a benchmark, inequality, ratio limit, or improper integral, and state why the comparison applies.

Alternating and

Sign changes can create convergence even when the corresponding positive-term series diverges. An alternating series has 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 applies when the magnitudes are eventually decreasing and approach zero:

b_n\text{ decreases eventually},\qquad \lim_{n\to\infty}b_n=0.

For example, the alternating harmonic series converges because $1/n$ decreases to zero:

\sum_{n=1}^{\infty}(-1)^{n-1}\frac1n.

Its absolute-value series is $\sum1/n$ , which diverges. Therefore, the alternating harmonic series is conditionally convergent rather than absolutely convergent.

When the applies, the error after the $N$ th partial sum satisfies

|S-S_N|\le b_{N+1}.

Thus, the first omitted term gives a direct error bound. For instance, after ten terms of the alternating harmonic series, the error is at most $1/11$ .

means that $\sum|a_n|$ converges. It always implies convergence of $\sum a_n$ . means that the original series converges while its absolute-value series diverges.

Takeaway: For alternating signs, inspect the positive magnitudes, verify both conditions, and distinguish from .

The and Test Selection

The is especially effective for factorials, powers, exponentials, and products. For $\sum a_n$ , calculate

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

The conclusions are:

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

For $a_n=n!/3^n$ ,

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

so $\sum n!/3^n$ diverges. For $a_n=2^n/n!$ ,

\left|\frac{a_{n+1}}{a_n}\right|=\frac{2}{n+1}\to0,

so $\sum2^n/n!$ converges absolutely.

For a power series

\sum_{n=0}^{\infty}c_n(x-a)^n,

the often identifies a radius of convergence $R$ . Typically, the test gives convergence for $|x-a|<R$ and divergence for $|x-a|>R$ . It does not settle the endpoints $|x-a|=R$ ; each endpoint requires a separate test.

A practical decision process is:

Check whether the terms approach zero.
Look for a geometric ratio.
Look for telescoping cancellation.
Check for alternating signs.
Compare with a p-series or .
Use the for a suitable positive, decreasing function.
Use the for factorials, powers, exponentials, or products.
If a test is inconclusive, choose another test rather than forcing a conclusion.

Takeaway: Choose a test that matches the algebraic structure, verify its hypotheses, and interpret its conclusion precisely.