A bound of the harmonic series of squares.

Question

I’ve been computing for some values of the sum and it seems that:

$$ \sum_{k=n+1}^\infty \frac{1}{k^2} < \frac{1}{n} $$

But I’m not sure how to prove it… I tried induction but it doesn’t work… is there any other way?

This has to do in specific with the solution of Durrett’s book of probability: exercise 2.2.4

Answer 1

Without using integral, there's the classical trick:

$$\frac{1}{k^2} < \frac{1}{k(k - 1)} = \frac{1}{k - 1} - \frac{1}{k}.$$

Summing up gives what you want.