How would you test for convergence with a series such as this using the alternating series test?
$1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-\frac{1}{7}...-\frac{1}{10}+\frac{1}{11}...+\frac{1}{15}-....$
I have attempted to find a summation equation for this by grouping in different ways in order to use the alternating series test but I can't find a way to group these in a logical way. Is there a grouping that would allow for analysis of this series using the alternating series test?
I hope I'm interpreting this correctly: you have one $+$ term, then two $-$, three $+$, four $-$, etc. The start and end of a run of $2k+1$ $+$ terms are $1/(2 k^2 + k + 1)$ and $1/(2 k^2 + 3 k + 1)$ and the start and end of the next run of $2k+2$ $-1$ terms are $-1/(2 k^2 + 3 k + 2)$ and $-1/(2 k^2 + 5 k + 3$. Thus the sum of these two runs is at least
$$ \dfrac{2k+1}{2 k^2 + 3k + 1} - \dfrac{2k+2}{2k^2+3k+2} = \frac{-k}{(2 k^2 + 3k + 1)(2k^2+3k+2)}$$ and at most
$$ \dfrac{2k+1}{2 k^2 +k + 1} - \dfrac{2k+2}{2k^2+5k+3} = \frac{-k}{(2 k^2 + 3k + 1)(2k^2+3k+2)} = \frac{6k+1}{(2k+3)(2k^2+k+1)}$$
Thus in any case it is bounded in absolute value by $C/k^2$ for some $C$, and thus the series converges.