Series Convergence and Divergence Practice Examples 5

Series Convergence and Divergence Practice Examples 5

We will now look at some more examples of applying the various convergence/divergence tests we have looked at so far to some series without being given what test to apply specifically.

More examples of evaluating series can be found on the following page:

Example 1

Does the series $\sum_{n=1}^{\infty} \frac{(-1)^n n^2}{n^2 + 5}$ converge or diverge?

We note that this is an alternating series, so let's try to apply the alternating series test. We first have that $\lim_{n \to \infty} \frac{n^2}{n^2 + 5} = 1$, and so we note that $\lim_{n \to \infty} \frac{(-1)^n n^2}{n^2 + 5} \neq 0$. Therefore we note that this series fails the alternating series test.

However we also note that since $\lim_{n \to \infty} \frac{(-1)^n n^2}{n^2 + 5} \neq 0$ we have that this series diverges by the divergence test.

Example 2

Does the series $\sum_{n=1}^{\infty} \frac{1}{1 + 10n}$ converge or diverge?

In this example we will apply the integral test. Let $f(x) = \frac{1}{1 + 10x}$, and we will evaluate the following integral:

(1)
\begin{align} \int_{1}^{\infty} \frac{1}{1 + 10x} \: dx \end{align}

So let $u = 1 + 10x$. Therefore $du = 10 \: dx$ and so $\frac{1}{10} du = dx$, and so:

(2)
\begin{align} \quad \int \frac{1}{1 + 10} \: dx = \frac{1}{10} \frac{1}{u} \: du = \frac{1}{10} \ln u + C = \frac{1}{10} \ln (1 + 10x) + C \end{align}

Continuing on we get:

(3)
\begin{align} \quad \int_{1}^{\infty} \frac{1}{1 + 10x} \: dx = \lim_{b \to \infty} \int_{1}^{b} \frac{1}{1 + 10x} \: dx = \lim_{b \to \infty} \left [ \frac{1}{10} \ln (1 + 10x) \right ]_{1}^{b} = \lim_{b \to \infty} \left [ \frac{1}{10} \ln (1 + 10b) \right ] - \left [ \frac{1}{10} \ln (11) \right ] = \infty \end{align}

Therefore by the integral test $\sum_{n=1}^{\infty} \frac{1}{1 + 10n}$ diverges.

Example 3

Does the series $\sum_{n=1}^{\infty} \frac{(2n)!}{(n!)^3}$ converge or diverge?

Applying the ratio test we have that:

(4)
\begin{align} \quad \lim_{n \to \infty} \frac{\frac{(2(n+1))!}{((n+1)!)^3}}{\frac{(2n)!}{(n!)^3}} = \lim_{n \to \infty} \frac{(2n+2)!(n!)^3}{(2n)!((n+1)!)^3} = \lim_{n \to \infty} \frac{(2n+1)(2n+2)(n!)^3}{((n+1)!)^3} = \lim_{n \to \infty} \frac{(2n+1)(2n+2)n!n!n!}{(n+1)!(n+1)!(n+1)!} = \lim_{n \to \infty} \frac{(2n+1)(2n+2)}{(n+1)^3} = 0 \end{align}

So by the ratio test $\sum_{n=1}^{\infty} \frac{(2n)!}{(n!)^3}$ converges.

Example 4

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

In this example we note that $\frac{n}{n^2 + 1} ≤ \frac{n}{n^2} = \frac{1}{n}$. However we have that $\sum_{n=1}^{\infty} \frac{1}{n}$ diverges, and so using this series to compare with won't give us any information. We will instead use the integral test. Let $f(x) = \frac{x}{x^2 + 1}$.

(5)
\begin{align} \int_{1}^{\infty} \frac{x}{x^2 + 1} \: dx \end{align}

Now let $u = x^2 + 1$ and so $du = 2x \: dx$ which implies that $\frac{1}{2} du = x \: dx$. Therefore:

(6)
\begin{align} \quad \int \frac{x}{x^2 + 1} \: dx = \frac{1}{2} \int \frac{1}{u} \: du = \frac{1}{2} \left [ \ln u \right ] + C = \frac{1}{2} \left [ \ln (x^2 + 1) \right ] + C \end{align}

Therefore:

(7)
\begin{align} \int_{1}^{\infty} \frac{x}{x^2 + 1} \: dx = \lim_{b \to \infty} \int_1^b \frac{x}{x^2 + 1} \: dx = \lim_{b \to \infty} \left ( \frac{1}{2} \left [ \ln (x^2 + 1) \right ] \right )_1^b = \lim_{b \to \infty} \frac{1}{2} \left [ \ln (b^2 + 1) \right ] - \frac{1}{2} \ln (2) = \infty \end{align}

And so by the integral test it follows that $\sum_{n=1}^{\infty} \frac{n}{n^2 + 1}$ diverges.

Example 5

Does the series $\sum_{n=1}^{\infty} \frac{6n^4 - 2n + \sqrt{n}}{2n^4 + 3n^3 + 4}$ converge or diverge?

We note that:

(8)
\begin{align} \lim_{n \to \infty} \frac{6n^4 - 2n + \sqrt{n}}{2n^4 + 3n^3 + 4} = \lim_{n \to \infty} \frac{6 - \frac{2}{n^3} + \frac{1}{n^{7/2}}}{2 + \frac{3}{n} + \frac{4}{n^4}} = 3 \end{align}

So by the divergence test we have that $\sum_{n=1}^{\infty} \frac{6n^4 - 2n + \sqrt{n}}{2n^4 + 3n^3 + 4}$ diverges.