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

Let $S_n(p) = \displaystyle \sum_{k=1}^n \dfrac1{k^p}$ and $S(p) = \lim_{n \to \infty} S_n(p)$. We then have
$$S_{2n+1}(p) = \sum_{k=1}^{2n+1} \dfrac1{k^p} = 1 + \sum_{k=1}^n \left(\dfrac1{(2k)^p} + \dfrac1{(2k+1)^p}\right) < 1 + \sum_{k=1}^n \dfrac2{(2k)^p} = 1 + \dfrac{S_n(p)}{2^{p-1}}$$
Letting $n \to \infty$, we get that
$$S(p) \leq 1 + \dfrac{S(p)}{2^{p-1}} \implies S(p) \leq \dfrac{2^{p-1}}{2^{p-1}-1}$$
