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.

Despite the fact that such solution is given in other posts (for example in this question about similar infinite series: Proof that $\sum_{1}^{\infty} \frac{1}{n^2} <2$ it is used as an auxiliary result in some answers) it might be useful to mention the solution using telescoping sum in a question asking about finite sum.
Let us look at the sum starting with $k=2$:
$$\sum\limits_{k=2}^n \frac1{k^2} \le \sum\limits_{k=2}^n \frac1{k(k-1)}.$$
After rewriting this as  $\frac1{k(k-1)}=\frac1{k-1}-\frac1k$we see that on the RHS we get a telescoping sum, where many terms will cancel out:
$$\sum_{k=2}^n \left(\frac1{k-1}-\frac1k\right)=(1-\frac12)+\left(\frac12-\frac13\right)+\dots+\left(\frac1{n-1}-\frac1n\right)=1-\frac1n$$
(We started with $k=2$ so that we do not get zero in the denominator in the expressions $\frac1{k-1}$ and $\frac1{k(k-1)}$.)
