Using mathematical induction to prove $\frac{1}1+\frac{1}4+\frac{1}9+\cdots+\frac{1}{n^2}<\frac{4n}{2n+1}$

Question

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.

Answer 1

Hint: What is $\frac{4(n+1)}{2(n+1)+1}-\frac{4n}{2n+1}$? (What is the corresponding difference on the other side of the inequality when you move from $n$ to $n+1$?)

Answer 2

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

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