Series

Summable Sequence

The sequence $\{a_n\}$ is called summable if the sequence $\{s_n\}$ converges, where

In this case, $\lim \ s_n$ is denoted by and is called the sum of the sequence $\{a_n\}.$

Cauchy Criterion

The sequence $\{a_n\}$ is summable if and only if
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Therefore, if $\{a_n\}$ is summable then

Divergence of

Grouping the series as
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we see that the sum of the terms in each ( ) is $\geq \frac 12$ ; thus $\{s_n\}$ is unbounded. Therefore the series $\{\frac 1n\}$ is not summable.

Geometric Series

The infinite series is called the geometric series. If then each individual term does not converge to so the series diverges. If , then it follows from the two equations

that

or
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Since , it follows that
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Convergence Criterion for Nonnegative Series

Boundedness Criterion

A nonnegative series $\{a_n\}$ is convergent if and only if the set of partial sums is bounded.

Comparison Test

1. Suppose that

Then if converges, so does

Proof. If

then

Since converges, $\{t_n\}$ is bounded. Therefore $\{s_n\}$ is bounded; consequently is convergent.

2. If , and , then converges if and only if converges.

Proof. Suppose that converges. Since , there exists some such that

Since the series converges, it follows that the series converges and this implies the convergence of the whole series which has only finitely many additional terms.

The converse follows at once since

The Ratio Test

Let for all , and suppose that
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(a) If then converges.

(b) If then the terms of do not approach so diverges.

Proof. (a) Suppose that . Fix with . The assumption implies that there exists some such that
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i.e.,

Thus

Since converges, it follows that converges. Therefore the whole series converges.

(b) Suppose that . Fix with . Then there exists some such that
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which means

This shows that the individual terms of $\{a_n\}$ do not approach , so $\{a_n\}$ is not summable.

The Root Test

Let $a_n\geq 0$ for all , and suppose that

(a) If then converges.

(b) If then the terms of do not approach so diverges.

Proof. (a) Suppose that . Fix with . The assumption

implies that there is some such that

i.e.,

Thus

Since the geometric series converges, it follows that converges. Therefore the whole series converges.

(b) Suppose that . Fix with . Then there exists some such that

which means

This shows that the individual terms of $\{a_n\}$ do not approach , so $\{a_n\}$ is not summable.

Convergence of the Exponential Series

Let be any positive number. The ratio
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converges to as approaches infinity. Hence it follows from the ratio test that the series converges for any positive number . Consequently we have
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Improper Integral

If the limit exists, it is denoted by , and called an improper integral.

Find if .
Show that does not exist.
Suppose that $f(x)\geq 0$ for $x\geq 0$ and that exists. Show that if for all $x\geq 0$ , then exists.

The improper integral is defined as . If both improper integrals and exist, then the improper integral exists and equals .

Note that does not exist although the limit exists.

Prove that if exists then exists and equals

There is another kind of improper integral in which the interval is bounded, but the function is unbounded. If is an unbounded function such that for any $\epsilon$ with is bounded on $[a+\epsilon ,b]$ and the limit
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exists, then the improper integral $\int_a^bf(x)dx$ is said to exist and is equal to Similarly, if is an unbounded function such that for any $\epsilon$ with is bounded on $[a,b-\epsilon ]$ and the limit
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exists, then the improper integral $\int_a^bf(x)dx$ is said to exist and is equal to

Integral Test

Suppose that is positive and decreasing on $[1,\infty )$ , and that for all . Then converges if and only if the improper integral
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Proof. The convergence of is equivalent to the convergence of the series
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Since is decreasing we have
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Therefore, if exists, then the series converges and so it follows from the comparison test that converges, and so converges.