# How to prove using induction that $\sum_{i = 1}^n \frac{1}{\sum_{n = 0}^i n} = \frac{2n}{n+1}$

Using Mathematical Induction I need to prove that

$\sum_{i = 1}^n \frac{1}{\sum_{n = 0}^i n} = \frac{2n}{n+1}$

As A first step I verified it for numbers 1 and 2 which worked

Secondly I simplified the LHS as:

$\frac{1}{\sum n} = \frac{1}{\frac {n(n+1)}{2}} = \frac{2}{n(n+1)}$

which converts my problem into

$\frac{2}{1 \cdot 2} + \frac{2}{2 \cdot 3} + \frac{2}{3 \cdot 4} ......... + \frac{2}{n \cdot (n+1)} = \frac{2n}{n+1}$

After which I cancel out 2 as the common factor from both sides. But at this stage I am stuck and am unable to go forward. What should I do next ?

This is known as a Telescoping Series. To solve these, you use the method of Partial Fractions, in which you solve the equation $\frac A{n+1}+\frac B n=\frac1{n(n+1)}$, where $A$ and $B$ are constants. This will allow you to split up the function into something like the form $(\frac1 1-\frac 1 2)+(\frac 1 2 -\frac 1 3)+(\frac 1 3-\frac 1 4)+...$, which you can regroup into $\frac1 1-(\frac 1 2+\frac 1 2 )-(\frac 1 3+\frac 1 3)-(\frac 1 4+...=1-0-0-...$. Notice that the end term will not cancel out.

If you just want an induction proof, just substitute $n+1$ for $n$ and look for the value you got for $n$ on both sides, subtract that from both sides, and prove that what remains on both sides is equal.

Straightforward. After simplifying your expression, proceed with your induction. $$\sum_{i=1}^{n+1} \dfrac{2}{i(i+1)} = \sum_{i=1}^{n} \dfrac{2}{i(i+1)} + \frac{2}{(n+1)(n+2)} = \frac{2n}{n+1} + \frac{2}{(n+1)(n+2)} = \frac{2n^2+4n+2}{(n+1)(n+2)} = \frac{2(n+1)}{n+2}$$

• +1 because this answer actually gives a proof for induction. My answer just glosses over it in the end. – AlgorithmsX Nov 20 '16 at 4:08

To do this by induction, I start at the OP's simplification

$\sum_{k=1}^n \frac1{k(k+1)} =\frac{n}{n+1}$.

This is true for $n=1$, where it says $\frac12 = \frac12$.

If it is true for $n$, then

$\begin{array}\\ \sum_{k=1}^{n+1} \frac1{k(k+1)} &=\sum_{k=1}^{n} \frac1{k(k+1)}+\frac1{(n+1)(n+2)}\\ &=\frac{n}{n+1}+\frac1{(n+1)(n+2)}\\ &=\frac{n(n+2)+1}{(n+1)(n+2)}\\ &=\frac{n^2+2n+1}{(n+1)(n+2)}\\ &=\frac{(n+1)^2}{(n+1)(n+2)}\\ &=\frac{n+1}{n+2}\\ \end{array}$

and it is true for $n+1$.

Therefore it is true for all $n$.