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

$$\sum_{n=1}^\infty{1\over n^2}<1+\sum_{n=2}^\infty{1\over n^2-{1\over4}}= 1+\sum_{n=2}^\infty\left({1\over n-{1\over2}}-{1\over n+{1\over2}}\right)=1+{2\over3}={5\over3}\ .$$
By the way: Since the left hand side is $={\pi^2\over6}$ you obtain from this the estimate $\pi^2<10$, which is not bad. In fact $\pi^2\doteq9.87$.
