Show that
$$\sum^n_{j=1}\frac{1}{j^2}\leq 2-\frac{1}{n}$$
$\forall n\geq 1$
Let us denote our proposition by $P_n$. Testing for $P_1$ this shows us that
$$\sum^1_{j=1}\frac{1}{1}\leq2-1$$
$\therefore{\text{true for $P_1$}}$
Now, assuming that it holds for some natural number $m$. $P_m$ is true, where $m$ has the same boundaries as $n$. I.e $m\geq1$. Now we want to show as $P_m$ is true $P_m\rightarrow P_{m+1}$ $$P_m:\sum^m_{j=1}\frac{1}{j^2}\leq2-\frac{1}{m}$$ $$P_{m+1}: \sum^{m+1}_{j=1}\frac{1}{j^2}\leq2-\frac{1}{m+1}$$
We can manipulate the series to prove that it holds for $P_{m+1}$ too.
rewriting the sum we get $$\sum^{m+1}_{j=1}\frac{1}{j^2}=\sum^{m}_{j=1}\frac{1}{j^2}+\frac{1}{(m+1)^2}$$
$$\leq2-\frac{1}{m}+\frac{1}{(m+1)^2}$$ $$=2-\frac{(m+1)^2-m}{m(m+1)^2}$$ $$\leq2-\frac{1}{m+1}-\frac{2m-1}{m(m+1)^2}$$
This is where I'm confused. I'm stuck with something that does not equals our hypothesis. Have I done something wrong? Cause how I see it, most of the tricks in induction is just being good at algebra.
