Proving using Mathematical Induction from my discrete math class This is a practice exercise from our class about proving inequalities using mathematical induction. I've been stuck on the last step for quite a while now.
This is the Question. "Prove that $\sum_{k=1}^n\frac{1}{k^2}=\frac{1}{1^2}+\frac{1}{2^2}+\cdots+\frac{1}{n^2}\le 2-\frac{1}{n}$ whenever $n$ is a positive integer."
This is my attempt to prove it.
Step 1: Base case $(n=1)$
$\sum_{k=1}^1\frac{1}{(1)^2}\le2-\frac{1}{1}$
$1\le1$
Step 2: Induction hypothesis
Assume that $\sum_{k=1}^n\frac{1}{k^2}\le 2-\frac{1}{n}$ is true for $n=c$
Then, $\sum_{k=1}^c\frac{1}{k^2}\le 2-\frac{1}{c}$
Step 3: prove that it is true for $n=c+1$
$\sum_{k=1}^{c+1}\frac{1}{k^2}$ $\le$ $2-\frac{1}{c}+\frac{1}{(c+1)^2}$
$\le 2+\frac{-(c+1)^2+c}{c(c+1)^2}$
$\le 2+\frac{-(c^2+2c+1)+c}{c(c+1)^2}$
$\le 2+\frac{-c^2-2c-1+c}{c(c+1)^2}$
$\le 2-\frac{c^2+c+1}{c(c+1)^2}$
$\le 2-\frac{c^2+c}{c(c+1)^2}-\frac{1}{c(c+1)^2}$
$\le 2-\frac{c(c+1)}{c(c+1)^2}-\frac{1}{c(c+1)^2}$
$\le 2-\frac{1}{c+1}-\frac{1}{c(c+1)^2}$
I'm Stuck on this step.

In step3, you added a new term $\frac{1}{c+1}$ (while $n=c+1$) to the both sides of the induction hypothesis ($\sum_{k=1}^{c}\frac{1}{k^2} \le 2-\frac{1}{c}$) and ended with:
$\sum_{k=1}^{c+1}\frac{1}{k^2} \le 2-\frac{1}{c}+\frac{1}{(c+1)^2} \le ...
\le 2-\frac{1}{c+1}-\frac{1}{c(c+1)^2}$
Because $n=c+1$ and $n>1$, then $c>0$, then $\frac{1}{c(c+1)^2}>0$. Now we have:
$\sum_{k=1}^{c+1}\frac{1}{k^2} \le ... \le 2-\frac{1}{c+1}-\frac{1}{c(c+1)^2} \le 2-\frac{1}{c+1}$
And you are done with the deduction.
