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.

Just for the fun of it, you can re-write your inequality as:
$$\frac{1}{n}-\frac{1}{n+1}>\frac{1}{(n+1)^2}.$$
This can be re-writen as:
$$\int_{n}^{n+1}\frac{1}{x^2}dx>\frac{1}{(n+1)^2},$$
or, if $f(x)=\dfrac{1}{x^2}$:
$$\int_{n}^{n+1}f(x)dx>f(n+1),$$
which is obvious if you consider that $f$ is strictly decreasing and positive on $(0,+\infty)$ and that the right-hand side of the inequality is the area of a rectangle with height $f(n+1)$ and a basis of $1$ - see the figure below.

