Help with the algebra in for this number theory proof? For all $n\geq 1$, prove with mathematical induction 
$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}\leq 2-\frac{1}{n}$
So far.. I have substituted 1 and saw that the statement is true and I have plugged in n+1 to show that the proof is true for all integers but I don't know how to go about the simplification.. right now I have 
LHS: $2-\frac{1}{k}+\frac{1}{(k+1)^2} \leq 2-\frac{1}{k+1}$
Should I try to find common denominators for the left? Step by step explanation please!

you simply have
$$2-\frac{1}{k}+\frac{1}{(k+1)^2} \\
=2-\{\frac{1}{k}-\frac{1}{(k+1)^2}\} \\
=2-\{\frac{(k+1)^2-k}{k(k+1)^2}\}\\
=2-\{\frac{k^2+k+1}{k(k+1)^2}\}\\
\leq 2-\frac{k(k+1)}{k(k+1)^2}$$
Since
$$k^2+k+1 \gt k^2+k\\
\frac{k^2+k+1}{k(k+1)^2} \gt \frac{k^2+k}{k(k+1)^2}\\
 -\frac{k^2+k+1}{k(k+1)^2} \le -\frac{k^2+k}{k(k+1)^2}$$
