Question

Find the sum of the series?

`1/1+1/(1+2)+1/(1+2+3)+...+1/(1+2+3+... 100)`

Answer 1

`200/101`

Explanation:

First, make use of the following fact:

`1+2+3+...+n=(n(n+1))/2`

This is also the formula for Triangular Numbers, which is the series of numbers {1, 3, 6, 10, ...}

Thus the series

`1/1+1/(1+2)+1/(1+2+3)+...+1/(1+2+3+...+100)`

becomes

`1/((1(1+1))/2)+1/((2(2+1))/2)+1/((3(3+1))/2)+...+1/((100(100+1))/2)`

If you multiply the numerators and denominators by 2, you get

`2/(1(1+1))+2/(2(2+1))+2/(3(3+1))+...+2/(100(100+1))`

Factorising out a 2, you get

`2*[1/(1(1+1))+1/(2(2+1))+1/(3(3+1))+...+1/(100(100+1))]`

This is actually a telescoping series. Observe that

`1/(n(n+1))=1/n-1/(n+1)`

So the series can be re-written as

`2*[(1/1-1/(1+1))+(1/2-1/(2+1))+(1/3-1/(3+1))+...+(1/100-1/(100+1))]`

And you may notice a nice cancellation occurring here:

`2*[1/1-cancel(1/(2))+cancel(1/2)-cancel(1/(3))+cancel(1/3)-cancel(1/4)+...+cancel(1/100)-1/(101)]`

So the answer is just `2*[1-1/101]=2*100/101=200/101`