<s>[INST] Prove: $\frac{1}{1^2} +\frac{1}{2^2} + \cdots + \frac{1}{n^2} + \cdots = \sum_{n=1}^\infty \frac{1}{n^2} < 2$ While I don't doubt that this question is covered somewhere else I can't seem to find it, or anything close enough to which I can springboard. I however am trying to prove $$\frac{1}{1^2} +\frac{1}{2^2}  + \cdots + \frac{1}{n^2} + \cdots = \sum_{n=1}^\infty \frac{1}{n^2} < 2$$ by induction.
I have seen it many times and proved it before but can't remember what it was I did.  I see that for the first two terms $n = 1, n=2$ I get:
for $n = 1$, $\frac{1}{1^2} = 1 < 2$
for $n = 2$, $\frac{1}{1^2} + \frac{1}{2^2} = \frac{5}{4} < 2$
Now I am stumped, I know I want to show this works for the $n+1$ term and am thinking, let the series $\sum_{n=1}^\infty \frac{1}{n^2} = A(n)$ Then look to show the series holds for $A(n+1)$ But $A(n+1) = A(n) + \frac{1}{(n+1)^2}$  But now what?  If I tried $A(n+1) - A(n) = \frac{1}{(n+1)^2}$ , but would have to show that this is less than $2 - A(n)$.  I am stuck.
Thanks for your thoughts, 
Brian
 [/INST] Hint Prove instead that
$$\frac{1}{1^2} +\frac{1}{2^2}  + \cdots + \frac{1}{n^2}  < 2-\frac{1}{n}$$
Interesting, this stronger exercise is an easy induction problem, while the weaker result you mentioned cannot be proven directly by induction for obvious reasons (LHS increases, RHS is constant < --- in the stronger version both sides are increasing, this issue is fixed).
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