Proving $\sum_{r=1}^{n}{\frac{1}{r^2}\leq2-\frac{1}{n}}$ by induction

Question

After checking $n=1$ I assumed the statement true for some $k$ and then tried with :$$\begin{align}\sum_{r=1}^{k+1}{\frac{1}{r^2}}&=\sum_{r=1}^{k}{\frac{1}{r^2}}+\frac{1}{(k+1)^2}\\&\leq2-\frac{1}{k}+\frac{1}{(k+1)^2}\end{align}$$

I wasn't sure how to keep going here though. The goal would be probably be $2-\frac{1}{k+1}$.

Answer 1

You are almost done. Note that $$\begin{align*} \frac{1}{k} - \frac{1}{(k+1)^2} \geqslant \frac{1}{k+1} &\Leftrightarrow \frac{1}{k}-\frac{1}{k+1} \geqslant \frac{1}{(k+1)^2} \\ &\Leftrightarrow \frac{1}{k(k+1)} \geqslant \frac{1}{(k+1)^2} \\ &\Leftrightarrow (k+1)^2 \geqslant k(k+1).\end{align*}$$ The final inequality is immediately seen to be true.

Answer 2

You can just check the inequality directly: $$\begin{align*} \frac{1}{(k+1)^2}-\frac{1}{k} &= \frac{-k^2-k-1}{k(k+1)^2} \\\ &\le \frac{-k^2-k}{k(k+1)^2} \\\ &= -\frac{1}{k+1} \end{align*}$$

Answer 3

\begin{align}&\frac1k-\frac{1}{(k+1)^2}-\frac{1}{k+1}\\ &=\frac{k^2+2k+1-k-k^2-k}{k(k+1)^2}\\ &=\frac{1}{k(k+1)^2}\geq0 \end{align} Therefore, \begin{align}\sum_{r=1}^{k+1}{\frac{1}{r^2}}&=\sum_{r=1}^{k}{\frac{1}{r^2}}+\frac{1}{(k+1)^2}\\&\leq2-\frac{1}{k}+\frac{1}{(k+1)^2}\\ &\leq2-\frac{1}{k}+\frac{1}{(k+1)^2}+\frac1k-\frac{1}{(k+1)^2}-\frac{1}{k+1}\\&=2-\frac{1}{k+1} \end{align}

Answer 4

$2 - {1\over k} + {1\over{(k+1)^2}}\le 2 - {1\over k} + {1\over{k(k+1)}} = 2 - {1\over{k+1}}$