<s>[INST] Proving $1+ \frac 14 + \frac 19 + \ldots + \frac 1{n^2} \lt 2 - \frac 1n $ with induction Prove $$1+ \frac 14 + \frac 19 + \ldots + \frac 1{n^2} \lt 2 - \frac 1n $$
What I've noticed is that it only holds $\forall n \in \mathbb Z_0 - \{1\}$ because it doesn't say in my textbook to satisty such particular conditions. 
$\underline{ n= 2:} \qquad $ $1 + \frac 14 \lt 2 - \frac 12$
$\underline{I.H.:} (n=k)\qquad $ $$\sum_{i=1}^k \frac 1{i^2} \lt 2- \frac 1k   $$
Now it comes 
$\underline{n= k+1:}$ $$\sum_{i=1}^k \frac 1{i^2} + \frac 1{(k+1)^2} \lt 2- \frac 1k + \frac1{(k+1)^2}$$
$$\sum_{i=1}^k \frac 1{(i+1)^2} \lt 2 - \frac {(k+1)^2 + k} {k(k+1)^2}= \frac{2k(k+1)^2 - (k+1)^2 + k }{k(k+1)^2} $$
Am I right working in the right direction and could someone give me a hint on how to proceed?
 [/INST] \begin{eqnarray*}  
\sum_{i=1}^{n} \frac{1}{i^2}+ \frac{1}{(n+1)^2}< 2-\frac{1}{n} +\frac{1}{(n+1)^2} 
= 2- \frac{n^2+n+\color{orange}{1}}{n(n+1)^2} 
< 2 - \frac{n^2+n}{n(n+1)^2} \\
\end{eqnarray*}
</s>