If $\Sigma_{k=1}^n \frac{1}{k(k+1)}= \frac{7}{8}$ then what is $n$ equal to? If $S_n=\Sigma_{k=1}^n \frac{1}{k(k+1)}= \frac{7}{8}$ then what is $n$ equal to?
So, the most obvious course of action in my mind is to find a closed form for the partial summations, but alas, this task eludes me. I started doing this by hand... like just adding up the fractions until I get to $\frac{7}{8}$ and got $n=7$. Surely there must be a better way. Help appreciated here! Thanks, I really appreciate it.

this is a telescopic series, this means that each term cancel part of other term, in your case you first need to apply partial fraction on the given expression:
$\sum_{k=1}^n \frac{1}{k(k+1)} =\sum_{k=1}^n \frac{1}{k} - \frac{1}{k+1} $
this means that the first terms are:
$  k=1: 1-\frac{1}{2} $
$  k=2: \frac{1}{2}-\frac{1}{3} $
$  k=3: \frac{1}{3}-\frac{1}{4} $
$ ...$
$  k=n: \frac{1}{n}-\frac{1}{n+1} $
now you can see that when you sum all the term untill the n-th term you are getting:
$ 1-\frac{1}{n+1} $
now solve $ 1-\frac{1}{n+1} = \frac{7}{8} $ to get $ x=7 $
