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.
 A: Note that $$\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$$
Are you able to take it from here? As suggested in comments, you should able to recognize a telescopic sum.
A: 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?
A: 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
$$1-\frac{1}{n+1} = \frac{7}{8} \implies n = 7$$
A: 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 $
