Convergence Tests

Section 9.4 Convergence Tests

Note 9.4.1.

The notation \(\displaystyle \sum_{n=1}^{\infty} a_n\) can also be represented generally by \(\sum a_n\text{.}\)

Subsection 9.4.1 Divergence Test

Consider a series \(\sum a_n\text{:}\)

if \(\displaystyle \lim_{n\to\infty} a_n \neq 0\text{,}\) then \(\sum a_n\) diverges
if \(\displaystyle \lim_{n\to\infty} a_n = 0\text{,}\) then the test is inconclusive and could converge or diverge

Another way of stating this is, if \(\sum a_n\) converges then \(\displaystyle \lim_{n\to\infty} a_n = 0\text{.}\)

Example 9.4.1.

Consider:

\begin{equation*} \sum_{n=1}^{\infty} \frac{n}{n+1} = \frac{1}{2} + \frac{2}{3} + \frac{3}{4} + \dots \end{equation*}

\begin{equation*} \lim_{n\to\infty} \frac{n}{n+1} = 1 \neq 0 \quad \boxed{\text{diverges}} \end{equation*}

Subsection 9.4.2 Algebraic Properties of Series

If \(\sum a_n\) and \(\sum b_n\) are convergent series then \(\sum (a_n \pm b_n) \) also coverge, further

\begin{equation*} \sum (a_n \pm b_n) = \sum a_n \pm \sum b_n \end{equation*}
If \(c\) is a nonzero constant then, \(\sum a_n\) and \(\sum (c a_n)\) both converge or both diverge. If they converge:

\begin{equation*} \sum c a_n = c(\sum a_n) \end{equation*}
Convergence or divergence is unaffected by deleting a finite number of terms from a series. In particular, fro any positive integer \(k\) the two series:

\begin{equation*} \sum_{n = 1}^{\infty} a_n = a_1 + a_2 + a_3 + \dots \end{equation*}

and

\begin{equation*} \sum_{k = 1}^{\infty} a_n = a_k + a_{k+1} + a_{k+2} + \dots \end{equation*}

both converge or both diverge.

Example 9.4.2.

Find the sum of the series:

\begin{equation*} \sum_{n-1}^{\infty} \left( \frac{3}{4^{n}} - \frac{2}{5^{n-1}} \right) = \left( \sum_{n-1}^{\infty} \frac{3}{4^{n}} \right) - \left( \sum_{n-1}^{\infty} \frac{2}{5^{n-1}} \right) \end{equation*}

\begin{align*} \sum_{n-1}^{\infty} \frac{3}{4^{n}} \amp = \frac{3}{4} + \frac{3}{16} + \frac{3}{64} + \dots \\ \amp = \frac{a}{1 - x} = \frac{\frac{3}{4}}{1 - \frac{1}{4}} \\ \amp = 1 \end{align*}

\begin{equation*} a = \frac{3}{4}, \quad x = \frac{1}{4}, \quad \mid x \mid \lt 1 \end{equation*}

\begin{equation*} \text{converges} \end{equation*}

\begin{align*} \sum_{n-1}^{\infty} \frac{2}{5^{n-1}} \amp = 2 + \frac{2}{5} + \frac{2}{15} + \dots \\ \amp = \frac{a}{1 - x} = \frac{2}{1 - \frac{1}{5}}\\ \amp = \frac{5}{2} \end{align*}

\begin{equation*} a = 2, \quad x = \frac{1}{5}, \quad \mid x \mid \lt 1 \end{equation*}

\begin{equation*} \text{converges} \end{equation*}

\begin{align*} \sum_{n-1}^{\infty} \left( \frac{3}{4^{n}} - \frac{2}{5^{n-1}} \right) \amp = 1 - \frac{5}{2}\\ \amp = \boxed{- \frac{3}{2}} \end{align*}

Subsection 9.4.3 Integral Test

\begin{equation*} \sum_{n = 1}^{\infty} a_n = a_1 + a_2 + a_3 + a_4 + \dots \end{equation*}

There are two possibilities:

if \(\displaystyle \int_0^{\infty} f(x) dx\) converges, then \(\displaystyle \sum a_n\) converges
if \(\displaystyle \int_0^{\infty} f(x) dx\) diverges, then \(\displaystyle \sum a_n\) diverges

Definition 9.4.4. Integral Test.

Let \(\displaystyle \sum a_n\) be a series with positive terms. If \(f\) is a function that is decreasing and continuous on the interval \([a,\infty)\) such that \(a_n = f(n)\) for all \(n \geq a\) then

\begin{equation*} \sum_{n = 1}^{\infty} a_n \quad \text{and} \quad \int_1^{\infty} f(x) dx \end{equation*}

both coverge or both diverge.

Example 9.4.3.

Determine if \(\displaystyle \sum_{n = 1}^{\infty} \dfrac{1}{n}\) converges or diverges.

Consider \(\displaystyle \int_1^{\infty} \dfrac{1}{x} dx\text{:}\)

\begin{align*} \int_1^{\infty} \dfrac{1}{x} dx \amp = \lim_{b\to\infty} \int_1^b \dfrac{1}{x} dx \\ \amp = \lim_{b\to\infty} \left( \ln \mid x \mid \bigg|_1^b \right) \\ \amp = \lim_{b\to\infty} \left( \ln(b) - \ln(1) \right)\\ \amp = \infty \end{align*}

\begin{equation*} \boxed{ \text{Since } \int_1^{\infty} \dfrac{1}{x} dx \text{ diverges, } \sum_{n = 1}^{\infty} \dfrac{1}{n} \text{ also diverges by the integral test.} } \end{equation*}