Question

How do you prove that sum of infinite series `1+1/4+1/9+...............` is less than two?

Answer 1

See explanation...

Explanation:

`sum_(n=1)^oo 1/n^2 = 1 + sum_(n=2)^oo 1/n^2 < 1 + sum_(n=2)^oo 1/(n(n-1))`

Then:

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

So:

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

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

`color(white)(sum_(n=2)^N 1/(n(n-1))) = 1+sum_(n=3)^N 1/(n-1) - sum_(n=2)^(N-1) 1/n - 1/N`

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

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

So:

`sum_(n=2)^oo 1/(n(n-1)) = lim_(N->oo) (1-1/N) = 1`

So:

`sum_(n=1)^oo 1/n^2 < 1 + sum_(n=2)^oo 1/(n(n-1)) = 1 + 1 = 2`