Question

How do you find the sum of an infinite non-geometric series?

Answer 1

There is no one way and it can be difficult.

Explanation:

Consider the series:

`1/2+1/6+1/12+1/20+...+1/(n(n+1))+...`

This is not a geometric series, but we can find its sum as follows:

`sum_(n=1)^N 1/(n(n+1)) = sum_(n=1)^N ((n+1)-n)/(n(n+1))`

`color(white)(sum_(n=1)^N 1/(n(n+1))) = sum_(n=1)^N (1/n-1/(n+1))`

`color(white)(sum_(n=1)^N 1/(n(n+1))) = sum_(n=1)^N 1/n- sum_(n=2)^(N+1) 1/n`

`color(white)(sum_(n=1)^N 1/(n(n+1))) = 1+color(red)(cancel(color(black)(sum_(n=2)^N 1/n)))- color(red)(cancel(color(black)(sum_(n=2)^N 1/n))) - 1/(N+1)`

`color(white)(sum_(n=1)^N 1/(n(n+1))) = 1-1/(N+1)`

So:

`sum_(n=1)^oo 1/(n(n+1)) = lim_(N->oo) sum_(n=1)^N 1/(n(n+1))`

`color(white)(sum_(n=1)^oo 1/(n(n+1))) = lim_(N->oo) (1-1/(N+1))`

`color(white)(sum_(n=1)^oo 1/(n(n+1))) = 1`

That wasn't too tricky, and sometimes you can use similar methods to calculate similar sums, but...

Consider the similar-looking:

`sum_(n=1)^oo 1/n^2 = 1+1/4+1/9+1/16+1/25+...`

Calculating this infinite sum was known as the Basel Problem, first posed in 1644 by Pietro Mengoli. It was not solved until 90 years later in 1734 by Leonhard Euler.

In fact:

`sum_(n=1)^oo 1/n^2 = pi^2/6`

but it is not particularly easy to prove.

See https://socratic.org/s/aHtqQrcq