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.

There is a certain name for this type of sum...
First we can use partial fractions method:
$$
\frac{1}{k(k+1)} = \frac{A}k+\frac{B}{k+1} \quad \implies
$$
$$
1 = Ak+B(k-1) \quad \forall k
$$
So if we set $k=1$ we obtain that $A=1$.
If we set $k=0$ then $B = -1$ this means that
$$
\frac{1}{k(k+1)} = \frac{1}{k}-\frac{1}{k+1}
$$
Now the sum:
$$
\sum_{k=1}^s \frac{1}{k(k+1)} = \sum_{k=1}^s  \frac{1}{k}-\frac{1}{k+1}
$$
But certain terms in this series cancel out...
For example if $s  =3$ then
$$
\sum_{k=1}^3  \frac{1}{k}-\frac{1}{k+1} =  \frac{1}{1}-\frac{1}{2} + \frac{1}{2}-\frac{1}{3}+ \frac{1}{3}-\frac{1}{4} = 1-\frac{1}{4}
$$
Do you see the pattern?
