Find the sum of the series?

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

1 Answer
May 19, 2018

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