1. ## Series

I have only done simpler series before so this is new to me. Grateful for help, Thanks!

Determine the value of M so that the given series converges.

2. Originally Posted by Duffman
I have only done simpler series before so this is new to me. Grateful for help, Thanks!

Determine the value of M so that the given series converges.
If you simplify

$\frac{n+1}{n-6} + \frac{n+1}{n+7} - 2 - \frac{M}{n} = \frac{42M-(M-85)n-(M-1)n^2}{n(n-6)(n+7)}$

LCT with $\sum_{n=7}^{\infty} \frac{1}{n}$ will show it diverges for all values of M except $M = 1$ and if so, then

$\frac{n+1}{n-6} + \frac{n+1}{n+7} - 2 - \frac{M}{n} = \frac{42+84n}{n(n-6)(n+7)}$

LCT with $\sum_{n=7}^{\infty} \frac{1}{n^2}$ will show it converges.

3. By the divergence theorem the limit of the inside term needs to be zero in order to have convergence.
Therefore,
$\lim \frac{42M-(M-85)n-(M-1)n^2}{n(n-6)(n+7)} = 0 \implies M = 1$.