How to solve this elementary induction proof: $\frac{1}{1^2}+ \cdots+\frac{1}{n^2}\le\ 2-\frac{1}{n}$? This is a seemingly simple induction question that has me confused about perhaps my understanding of how to apply induction
the question;
$$\frac{1}{1^2}+ \cdots+\frac{1}{n^2}\ \le\ 2-\frac{1}{n},\ \forall\ n \ge1.$$
this true for $n=1$, so assume the expression is true for $n\le k$. which produces the expression, 
$$\frac{1}{1^2} + \cdots + \frac{1}{k^2} \le\ 2-\frac{1}{k}.$$  now to show the expression is true for $k+1$,
$$\frac{1}{1^2}+\cdots+ \frac{1}{k^2} + \frac{1}{(k+1)^2} \le\ 2-\frac{1}{k}+\frac{1}{(k+1)^2}.$$
this the part I am troubled by, because after some mathemagical algebraic massaging, I should be able to equate,
$$2-\frac{1}{k}+\frac{1}{(k+1)^2}=2-\frac{1}{(k+1)},$$
which would prove the expression is true for $k+1$ and I'd be done. right? but these two are not equivalent for even $k=1$, because setting $k=1$ you wind up with $\frac{5}{4}=\frac{3}{2}$, so somewhere i am slipping up and I'm not sure how else to show this if someone has some insight into this induction that I'm not getting. thanks.

What you really need is $2 − \frac{1}{k} + \frac{1}{(k+1)^2} \leq 2 − \frac{1}{(k+1)}$,
