Abel's Test for Convergence of Series of Real Numbers Examples 1

Recall from Abel's Test for Convergence of Series of Real Numbers page the following test for convergence/divergence of a geometric series:

Abel's Test for Convergence of Series of Real Numbers

Let $(a_n)_{n=1}^{\infty}$ and $(b_n)_{n=1}^{\infty}$ be sequences of real numbers.

If $\displaystyle{\sum_{n=1}^{\infty} a_n}$ converges, $(b_n)_{n=1}^{\infty}$ is monotonic, and $(b_n)_{n=1}^{\infty}$ converges, then we conclude that:

• $\displaystyle{\sum_{n=1}^{\infty} a_nb_n}$ converges.

Let's now look at some examples of using Abel's test.

Example 1

Show that the series $\displaystyle{\sum_{n=1}^{\infty} \frac{n^3n! \cos \left ( \frac{1}{n^2} \right )}{e^n(n+2)!}}$ converges.

Let $\displaystyle{(a_n)_{n=1}^{\infty} = \left (\frac{n^3n!}{e^n(n+2)!} \right )_{n=1}^{\infty}}$ and let $\displaystyle{(b_n)_{n=1}^{\infty} = \left ( \cos \left ( \frac{1}{n^2} \right ) \right )_{n=1}^{\infty}}$. Using the ratio test and we see that:

(1)

\begin{align} \quad \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \left ( \frac{(n+1)^3(n+1)!}{e^{n+1}(n+3)!} \frac{e^n(n+2)!}{n^3n!} \right ) = \lim_{n \to \infty} \frac{(n+1)^4}{en^3(n+3)} = \frac{1}{e} \end{align}

So by the ratio test we have that the series $\displaystyle{\sum_{n=1}^{\infty} a_n}$ converges. Furthermore, the sequence $(b_n)_{n=1}^{\infty}$ is decreasing and converges to $1$. So by Abel's test, $\displaystyle{\sum_{n=1}^{\infty} \frac{n^3n! \cos \left ( \frac{1}{n^2} \right )}{e^n(n+2)!}}$ converges.

Example 2

Show that the series $\displaystyle{\sum_{n=1}^{\infty} \left ( \frac{n^2 + 3n + 1}{n^4 + 2n^2} \cdot \sum_{k=1}^{n} \frac{2}{k^2} \right )}$ converges.

Let $\displaystyle{(a_n)_{n=1}^{\infty} = \left ( \frac{n^2 + 3n + 1}{n^4 + 2n^2} \right )_{n=1}^{\infty}}$ and let $\displaystyle{(b_n)_{n=1}^{\infty} = \left ( \sum_{k=1}^{n} \frac{2}{k^2} \right )_{n=1}^{\infty}}$.

Notice that for sufficiently large $n$ that:

(2)

\begin{align} \quad \frac{n^2 + 3n + 1}{n^4 + 2n^2} \approx \frac{n^2}{n^4} = \frac{1}{n^2} \end{align}

The series $\displaystyle{\sum_{n=1}^{\infty} \frac{1}{n^2}}$ converges, and using the limit comparison test with this series we see that:

(3)

\begin{align} \quad L = \lim_{n \to \infty} \left ( \frac{n^2 + 3n + 1}{n^4 + 2n^2} \cdot \frac{n^2}{1} \right ) = \lim_{n \to \infty} \frac{n^4 + 3n^3 + n^2}{n^4 + 2n^2} = 1 \end{align}

So by the limit comparison test, since $0 < L = 1 < \infty$ we have that $\displaystyle{\sum_{n=1}^{\infty}\frac{n^2 + 3n + 1}{n^4 + 2n^2}}$ converges.

Now look at the sequence $(b_n)_{n=1}^{\infty}$. Notice that this is simply the sequence of partial sums to the sequences $\displaystyle{\sum_{n=1}^{\infty} \frac{2}{n^2}}$ which converges. So $(b_n)_{n=1}^{\infty}$ converges and is clearly monotonic.

So, by Abel's test we conclude that $\displaystyle{\sum_{n=1}^{\infty} \left ( \frac{n^2 + 3n + 1}{n^4 + 2n^2} \cdot \sum_{k=1}^{n} \frac{2}{k^2} \right )}$ converges.