Reciprocal squares sum inequality

Question

What is the easiest (preferably inductional) way without approximation of the sum_ to prove the following inequality: $\frac{1}{1^2}+\frac{1}{2^2} + \ldots +\frac{1}{n^2} \le 2 - \frac{1}{n}$

This is a duplicate for sure - I saw the same question at most three days ago. — Jack D'Aurizio, Feb 23 '15 at 20:29

score 4 · Accepted Answer · answered Feb 23 '15 at 18:49

4

hint: $\dfrac{1}{n^2} \leq \dfrac{1}{n(n-1)} = \dfrac{1}{n-1} - \dfrac{1}{n}$

answered Feb 23 '15 at 18:49

DeepSea

77,651

score 0 · Answer 2 · answered Feb 23 '15 at 18:57

0

$$\sum_{k=1}^n\frac{1}{k^2}\ \le \sup_{n\ \in\ \Bbb N - \{0,1,2\}}\{\sum_{k=1}^n\frac{1}{k^2}\} = \sum_{k=1}^{\infty}\frac{1}{k^2} = \frac{\pi^2}{6} \le 2 - \frac{1}{n}, \forall \ n\ge 3$$

Moreover for $n \in \{1,2\}$ the property holds.

answered Feb 23 '15 at 18:57

An explanation from the down voter would be appreciated. – Feb 23 '15 at 20:27

Reciprocal squares sum inequality

2 Answers2