The Comparison Test for Positive Series of Real Numbers Examples 1

Recall from The Comparison Test for Positive Series of Real Numbers page the following test for convergence/divergence of a geometric series:

The Comparison Test for Positive Series of Real Numbers

Let $\displaystyle{\sum_{n=1}^{\infty} a_n}$ and $\displaystyle{\sum_{n=1}^{\infty} b_n}$ be positive series and suppose that there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then $a_n \leq b_n$.

a) If we have that the series $\displaystyle{\sum_{n=1}^{\infty} b_n}$ converges, then we conclude that:

• The series $\displaystyle{\sum_{n=1}^{\infty} a_n}$ converges.

b) If instead we have that the series $\displaystyle{\sum_{n=1}^{\infty} a_n}$ diverges, then we conclude that:

• The series $\displaystyle{\sum_{n=1}^{\infty} b_n}$ diverges.

We will now look at some examples of applying the comparison test.

Example 1

Determine whether $\displaystyle{\sum_{n=1}^{\infty} \frac{2n}{n^3 + ne^n + n}}$ converges or diverges.

We first simplify the general term of the series:

(1)

\begin{align} \quad \sum_{n=1}^{\infty} \frac{2n}{n^3 + ne^n + n} = \sum_{n=1}^{\infty} \frac{2}{n^2 + e^n + 1} \end{align}

Notice that for all $n \in \mathbb{N}$ we have that $n^2 < n^2 + e^n + 1$. Therefore:

(2)

\begin{align} \quad \frac{2}{n^2} > \frac{2}{n^2 + e^n + 1} \end{align}

The series $\displaystyle{\sum_{n=1}^{\infty} \frac{2}{n^2}}$ converges, and so by comparison, $\displaystyle{\sum_{n=1}^{\infty} \frac{2}{n^2 + e^n + 1}}$ also converges.

Example 2

Determine whether $\displaystyle{\sum_{n=1}^{\infty} \frac{e^n}{n + 2}}$ converges or diverges.

Notice that for all $n \in \mathbb{N}$ we have that $n < n + 2$. So:

(3)

\begin{align} \quad \frac{1}{n} > \frac{1}{n+2} \end{align}

For $n$ sufficiently large we we will have that $e^n > n + 2$ since exponential functions grow at a rate greater than that of linear functions. So, for $n$ sufficiently large:

(4)

\begin{align} \quad \frac{e^n}{n+2} > \frac{n+2}{n+2} = 1 \end{align}

Clearly $\displaystyle{\sum_{n=1}^{\infty} 1}$ diverges (since $\displaystyle{\lim_{n \to \infty} 1 \neq 0}$) by comparison, $\displaystyle{\sum_{n=1}^{\infty} \frac{e^n}{n + 2}}$ also diverges. $\blacksquare$