# Wolfram Alpha and Comparison Test, and Alternating Series Test

## The Series

$$\sum_\limits{n=1}^\infty (-1)^{n+1} \frac{4}{n+6}=\frac{4}7-\frac{4}8+\frac{4}9 …$$

## My Question

My first thought was to test for absolute convergence, and then I ended up using the alternating series test to finish the problem of but I still had some questions in my mind. Was the absolute convergence test necessary for it to be used? Secondly, would going to the alternating series test first be the optimal route for solving this problem? Moreover, this is a question of methodology.

## My Work

\begin{align} \sum_\limits{n=1}^\infty (-1)^{n+1} \frac{4}{n+6} &= \sum_\limits{n=1}^\infty a_n && \mathbf{Given} \\ \sum_\limits{n=1}^\infty \vert a_n \vert &= \sum_\limits{n=1}^\infty \frac{4}{n+6} && \mathbf{Absolute \ Convergence \ Test} \end{align}

After much thought at this point I decided to do the Comparison Test in order to make due with the problem of absolute convergence.

\begin{align} \because \sum_\limits{n = 1}^\infty \frac{4}{n+6} &\le \sum_\limits{n=1}^\infty \frac{4}{n} && \textbf{Comparison Test} \end{align} \begin{align} \therefore \textbf{Using \mathbf p-series, both series are divergent} \end{align}

Thus leading me to conclude that the series was not absolutely convergent, from here I went to use the alternating series to ultimately test if it was conditionally divergent.

\begin{align} \mathbf{Alternating\ Series \ Test} \\ a_{n+1} \le a_n \mathbf{for \ all \ n} \ \ \ \ \mathbf{True} \\ \lim_{n \to \infty} a_n &= \lim_{n \to \infty} (-1)^{n+1} \frac{4}{n+6}=0 \end{align} Therefore the series was conditionally convergent.

• Yes, the series is conditionally convergent by the tests you have used. – Kavi Rama Murthy Dec 23 '19 at 23:59
• It's $4\left(\ln2-1+\dfrac12-\dfrac13+\dfrac14-\dfrac15+\dfrac16\right)$ – J. W. Tanner Dec 24 '19 at 0:01

As a correction to your absolute comparison test, you've shown that the series is $$\le\infty$$, which is not helpful at all. In order to make a conclusion, you need to show that:
$$\sum_n|a_n|\ge\infty\\\text{or}\\\sum_n|a_n|\le S<\infty$$