Proof of an inequality by induction Prove using induction that
$$\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots+ \frac{1}{n^2} \le 2-\frac{1}{n}$$
for all positive whole numbers $n$.
I began by showing that it is true for $n=1$
I then assumed that it is true for $n=p$
$$\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots+ \frac{1}{p^2} = \sum_{k=1}^p \frac{1}{k^2} \le 2-\frac{1}{p}$$
I now want to show that it is true for $n=p+1$
$$\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots+ \frac{1}{p^2} + \frac{1}{(p+1)^2}= \sum_{k=1}^{p+1} \frac{1}{k^2} $$
If I add $\frac{1}{(p+1)^2}$ to $\sum_{k=1}^{p} \frac{1}{k^2}$, I will then get
$$\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots+ \frac{1}{p^2} + \frac{1}{(p+1)^2} \le 2-\frac{1}{p} + \frac{1}{(p+1)^2}$$
If this is true then 
$$2-\frac{1}{(p+1)}=2-\frac{1}{p} + \frac{1}{(p+1)^2}$$
or
$$-\frac{1}{(p+1)}=-\frac{1}{p} + \frac{1}{(p+1)^2}$$
$$0= \frac{1}{(p+1)} -\frac{1}{p} + \frac{1}{(p+1)^2}$$
$$0= \frac{p(p+1)}{p(p+1)^2} -\frac{(p+1)^2}{p(p+1)^2} + \frac{p}{p(p+1)^2}$$
$$0= \frac{p(p+1)}{p(p+1)^2} -\frac{(p+1)^2}{p(p+1)^2} + \frac{p}{p(p+1)^2}$$
$$0= \frac{p(p+1)-(p+1)^2 + p}{p(p+1)^2}$$
$$0= \frac{(p^2 + p) - (p^2 + 2p + 1) + p}{p(p+1)^2}$$
$$0= \frac{-1}{p(p+1)^2}$$
This is invalid. I am not sure where I have made a mistake but I think it is
$$2-\frac{1}{(p+1)}=2-\frac{1}{p} + \frac{1}{(p+1)^2}$$
It then must be that
$$-\frac{1}{(p+1)} < -\frac{1}{p} + \frac{1}{(p+1)^2}$$
$$0 \le \frac{1}{(p+1)} - \frac{1}{p} + \frac{1}{(p+1)^2}$$
$$ 0< \frac{-1}{p(p+1)^2}$$
which is true for all positive whole numbers $p$.
I am pretty sure it is proved now but I would be happy if someone can confirm this. 

I think the following induction a bit of better.
$$\sum_{k=1}^n\frac{1}{k^2}=1+\sum_{k=2}^n\frac{1}{k^2}<1+\sum_{k=2}^n\frac{1}{k(k-1)}=1+\sum_{k=2}^n\left(\frac{1}{k-1}-\frac{1}{k}\right)=1+1-\frac{1}{n}=2-\frac{1}{n}.$$
