Using mathematical induction to prove $\frac{1}1+\frac{1}4+\frac{1}9+\cdots+\frac{1}{n^2}<\frac{4n}{2n+1}$ This induction problem is giving me a pretty hard time: 

$$\frac{1}1+\frac{1}4+\frac{1}9+\cdots+\frac{1}{n^2}<\frac{4n}{2n+1}$$

I am struggling because my math teacher explained us that in this case ($n^2$) when we prove that $n+1$ satisfies the property we have to write it like this: $$ \frac{1}1+\frac{1}4+\frac{1}9+\cdots+  \frac{1}{n^2} +\frac{1}{(n+1)^2} < \frac{4(n+1)}{(2(k+1)+1)}$$ 
and I probably got lost in the process.

$$\frac{1}1+\frac{1}4+\frac{1}9+\ldots+\frac{1}{n^2}<\frac{4n}{2n+1}$$
fist proof for one:  $1< \frac{4}{3}$,
for k-> $$\frac{1}1+\frac{1}4+\frac{1}9+\ldots+\frac{1}{k^2}<\frac{4k}{2k+1}$$
for k+1-> $$\frac{1}1+\frac{1}4+\frac{1}9+\ldots+\frac{1}{n^2}+\frac{1}{(k+1)^2}<\frac{4k+4}{2k+3}$$
Now, 
$$\frac{4k}{2k+1}+\frac{1}{(k+1)^2}<\frac{4k+4}{2k+3}$$
if we demostrate that:
$$-\frac{4k}{2k+1}-\frac{1}{(k+1)^2}+\frac{4k+4}{2k+3}>0$$
then it is demonstrate
