11.4: Infinite series with positive terms

There are several tests for the convergence or divergence of infinite series with all positive terms. We consider two.

• Integral test
• Comparison test

Integral test

If $f$ is a function with $f(x)$ a decreasing continuous function defined for all numbers $x \geq k$, then the infinite series

$\sum_{n=k}^\infty f(n)$ converges if and only if the integral $\int_1^\infty f(x) dx$ converges.

Example

Use the integral test to determine whether or not $\sum_{n=1}^\infty \frac{1}{n}$ converges.

Solution

Indeed, it does not as

$$\begin{eqnarray*} \int_{1}^\infty \frac{dx}{x} & = & \lim_{r \to \infty} \int_1^... ...1}^{x=r} \\ & = & \lim_{r \to \infty} \ln(r) \\ & = & \infty \end{eqnarray*}$$

Another example

Does the series $\sum_{n=1}^\infty \frac{n}{e^n}$ converge?

Solution

Consider $f(x) = x e^{-x}$. We compute $f'(x) = (1 - x)e^{-x}$ which is negative for all $x > 1$. Thus, $f$ is decreasing.

We compute using integration by parts with $u = x$ so that $du = dx$ and $dv = e^{-x}$ so that $v = -e^{-x}$,

$$\begin{eqnarray*} \int_1^\infty x e^{-x} dx & = & \lim_{r \to \infty} (-x e^{-x}... ...to \infty} -(r + 1) e^{-r} + \frac{2}{e} \\ & = & \frac{2}{e} \end{eqnarray*}$$

Hence, $\sum_{n=1}^\infty \frac{n}{e^n}$ converges.

Comparison tests

If $a_1, a_2, a_3, \ldots$ and $b_1, b_2, b_3, \ldots$ are two sequences of positive numbers for which $a_i \leq b_i$ for every $i$, then

$$\text{if } \sum_{n=1}^\infty a_n \text{ diverges, so does } \sum_{n=1}^\infty b_n$$

while

$$\text{if } \sum_{n=1}^\infty b_n \text{ converges, so does } \sum_{n=1}^\infty a_n$$

moreover,

$0 \leq \sum_{n=1}^\infty a_n \leq \sum_{n=1}^\infty b_n$.

Examples

Does the series $\sum_{n=1}^\infty \frac{1}{n 2^n}$ converge?

Solution

Yes: $0 < \frac{1}{n 2^n} \leq \frac{1}{2^n}$ for every $n$. We know $\sum_{n=1}^\infty \frac{1}{2^n} = 1$. Hence, $\sum_{n=1}^\infty \frac{1}{n 2^n}$ converges and is at most $1$.