<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] $$\begin{array}{rcl}
\displaystyle \sum_{k=\color{red}{2}}^n \dfrac1{k^2}
&<& \displaystyle \sum_{k=2}^n \dfrac1{k(k-1)} \\
&=& \displaystyle \sum_{k=2}^n \left(\dfrac1{k-1} - \dfrac1k\right) \\
&=& \displaystyle \frac11-\frac12+\frac12-\frac13+\cdots+\frac1{n-1}-\frac1n \\
&=& \displaystyle 1-\frac1n \\
\displaystyle \sum_{k=\color{red}{1}}^n \dfrac1{k^2}
&<& \displaystyle \frac1{1^2}+1-\frac1n \\
&=& \displaystyle 2-\frac1n
\end{array}$$
Use this as a basis of your induction.
</s>