# How do you test the alternating series Sigma (-1)^(n+1)n/(10n+5) from n is [1,oo) for convergence?

Apr 16, 2018

Diverges by the Divergence Test.

#### Explanation:

The Alternating Series Test tells us that if we have a series $\sum {\left(- 1\right)}^{n} {b}_{n}$, where ${b}_{n}$ is a sequence of positive terms, the series converges if

a) ${b}_{n} \ge {b}_{n + 1}$, IE, the sequence is ultimately decreasing for all $n .$

b) ${\lim}_{n \to \infty} {b}_{n} = 0$

We should note that we don't need to have ${\left(- 1\right)}^{n + 1}$ -- any term that causes alternating signs, such as $\cos \left(n \pi\right) , {\left(- 1\right)}^{n - 1} , {\left(- 1\right)}^{n + 1}$, is okay.

Here, we see ${b}_{n} = \frac{n}{10 n + 5}$. Taking the limit,

${\lim}_{n \to \infty} \frac{n}{10 n + 5} = \frac{1}{10} \ne 0$ -- the Alternating Series Test is inconclusive here.

However, take the limit of the overall sequence:

${\lim}_{n \to \infty} {\left(- 1\right)}^{n + 1} \frac{n}{10 n + 5} \ne 0$ -- we can say this because although the limit does not truly exist, we can convince ourselves that it alternates signs and gets closer to $\frac{1}{10} ,$ causing divergence by the Divergence Test.