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 simple telescoping series trick
$$\sum_{k=1}^n \frac{1}{k(k+1)} = \frac{1}{k}-\frac{1}{k+1} \\ $$
By choosing values of k for $k = 1, \cdots, n$ then we have
$$\begin{align}k&=1 \implies 1-\frac{1}{2} \\ k&=2 \implies \frac{1}{2} - \frac{1}{3} \\\vdots \\ k&= n-1 \implies \frac{1}{n-1}- \frac{1}{n} \\ k&=n \implies \frac{1}{n} - \frac{1}{n+1} \end{align}$$
Everything cancels except for $1 -\frac{1}{n+1}$, therefore we have that
$$1-\frac{1}{n+1} = \frac{7}{8} \implies n = 7$$
