Show that $\sum_{i=1}^n \frac{1}{i^2} \le 2 - \frac{1}{n}$ So I am able to calculate the given problem and prove $P(K) \implies P(k + 1)$; it's been sometime since I did proofs
and I perform my steps I get what Wolfram Alpha shows as an alternate solution.
Any help is greatly appreciated
The problem is the following:
Show that 
$$\sum_{i=1}^n  \frac{1}{i^2} \le 2 - \frac{1}{n}$$
What I have:
$P(1)$:
$$\sum_{i=1}^n  \frac{1}{i^2} \le 2 - \frac{1}{n}$$
Replace n with 1
$$\frac{1}{1} \le 2 - \frac{1}{1}$$
Conclusion
$$1 \le 1$$
Prove:
$$\sum_{i=1}^{k+1}  \frac{1}{i^2} \le 2 - \frac{1}{k+1}$$
P(K) Assume 
  $$\sum_{i=1}^k  \frac{1}{i^2} \le 2 - \frac{1}{k}$$
$P(K) \implies P(k + 1)$
Performed Steps:
Working the LHS to match RHS
$$2 - \frac{1}{k} + \frac{1}{(k+1)^2}$$
Edit: Fixed error on regrouping
$$2 - \left[\frac{1}{k} - \frac{1}{(k+1)^2}\right]$$
Work the fractions
$$2 - \left[\frac{1}{k} \frac{(k+1)^2}{(k+1)^2} - \frac{1}{(k+1)^2} \frac{k}{k} \right]$$
$$2 - \left[\frac{(k+1)^2 - k}{k(k+1)^2} \right]$$
$$2 - \left[\frac{k^2 + 2k + 1 - k}{k(k+1)^2} \right]$$
$$2 - \left[\frac{k^2 + k + 1}{k(k+1)^2} \right]$$
$$2 - \left[\frac{k(k+1) + 1}{k(k+1)^2} \right]$$
$$2 - \left[\frac{k(k+1)}{k(k+1)^2} + \frac{1}{k(k+1)^2} \right]$$
$$2 - \frac{1}{(k+1)} - \frac{1}{k(k+1)^2}$$
EDIT: I fixed my mistake of my regrouping and signs; had completely missed the regrouping.
This is the final step I got to.  I am hung on where to go from here.  The answers given have been really helpful and I'm happy with them. I'd just like to know the mistake I made or next step I should take.
$$\sum_{i=1}^{k+1}  \frac{1}{i^2} \le 2 - \frac{1}{k+1}$$
Thanks for the help

You can approximate the sum by an integral to obtain the inequality
$$
\sum_{i=1}^n\frac1{i^2}=1+\sum_{i=2}^n\frac1{i^2}\le1+\int_1^n\frac1{x^2}\mathrm dx=2-\frac1n.
$$
