Prove that for every positive integer $n$, $1/1^2+1/2^2+1/3^2+\cdots+1/n^2\le2-1/n$ Base case: n=1. $1/1\le 2-1/1$. So the base case holds.
Let $n=k\ge1$ and assume 
$$1/1^2+1/2^2+1/3^2+\cdots+1/k^2\le 2-1/k$$
We want to prove this for $k+1$, i.e.
$$(1/1^2+1/2^2+1/3^2+\cdots+1/k^2)+1/(k+1)^2\le 2-\frac{1}{k+1}$$
This is where I get stuck. Any help appreciated.
 A: For your induction step, all you need is
$$
2-\frac{1}{k}+\frac{1}{(k+1)^2}\leq 2-\frac{1}{k+1}.
$$
That's equivalent to
$$
\frac{1}{k}-\frac{1}{(k+1)^2}-\frac{1}{k+1}\geq 0
$$
i.e.
$$
\frac{(k+1)^2-k-k(k+1)}{k(k+1)^2}=\frac{1}{k(k+1)^2}\geq 0.
$$
So it holds.
A: $\displaystyle \frac{1}{i^2}< \frac{1}{i(i-1)}=\frac{1}{i-1}-\frac{1}{i}$,for $i\ge 2$
So we have $\displaystyle 1+\sum_{i=2}^{k}\frac{1}{i^2}\le 1+\sum_{i=2}^{k}(\frac{1}{i-1}-\frac{1}{i})=2-\frac{1}{k}$
I think this is better than induction.
A: Let's assume it's true for $k$.
Then $\sum_{i=1}^k\dfrac{1}{i^2}\leq2-\dfrac 1 k$
Let's try $k+1$.
$\sum_{i=1}^{k+1}\dfrac{1}{i^2}=\sum_{i=1}^k\dfrac{1}{i^2}+\dfrac{1}{(k+1)^2}\leq2-\dfrac 1 k+\dfrac{1}{(k+1)^2}$
As $\dfrac{1}{(k+1)^2}-\dfrac 1 k<-\dfrac{1}{k+1}$ ,
$\sum_{i=1}^{k+1}\dfrac{1}{i^2}\leq2-\dfrac{1}{(k+1)^2}$
which shows that it is true for $k+1$.
A: I will give only a hint. Where you get stuck, you have by induction hypothesis that
$$1/1^2+\ldots+1/k^2\leq 2-1/k$$
Subtituting this in
$$1/1^2+\ldots+1/k^2+1/(k+1)^2$$
gives you
$$1/1^2+\ldots+1/k^2+1/(k+1)^2\leq 2-1/k+1/(k+1)^2$$
And then, you have only to prove
$$2-1/k+1/(k+1)^2\leq 2-1/(k+1)$$
Here, I think you will be able to rewrite the inequality in order to prove it.
