Proving inequality for induction proof: $\frac1{(n+1)^2} + \frac1{n+1} < \frac1n$ In an induction proof for $\sum_{k = 1}^n \frac{1}{k^2} \leq 2 - \frac{1}{n}$ (for $n \geq 1$), I was required to prove the inequality $\frac{1}{(n+1)^2} + \frac{1}{n+1} < \frac{1}{n}$.
This is my attempt:
$$ \begin{align} \frac{1}{(n+1)^2} + \frac{1}{n+1} &= \frac{1+(n+1)}{(n+1)^2} \\ 
                                                   &=\frac{n+2}{(n+1)^2} \\
                                                   &=\frac{n+2}{n^2 + 2n + 1} \\
                                                   &<\frac{n+2}{n^2 + 2n} (\text{since  } n \geq 1)\\
                                                   &=\frac{n+2}{n(n+2)}\\
                                                   &=\frac{1}{n} \end{align} $$
I am just wondering if there is a simpler way of doing this.

To prove the inequality you were required to prove, note
$\dfrac1{n+1}+\dfrac1{(n+1)^2}<\dfrac1{n+1}+\dfrac1{(n+1)^2}+\dfrac1{(n+1)^3}+...=\dfrac{\dfrac1{1+n}}{1-\dfrac1{n+1}}=\dfrac1n$
