Prove $\frac 12+\frac 23+\frac 34+\cdots+\frac n{n+1} \lt \frac {n^2}{n+1}$ using induction for $n \ge 2$

Question

Prove $\frac 12+\frac 23+\frac 34+\cdots+\frac n{n+1} \lt \frac {n^2}{n+1}$ using induction for $n \ge 2$.

It seems almost impossible for me to prove this. I know that $P(n)$ is true for $n = 2$. So all I need to do is prove that for $n+1$ this is true.

Since $P(k)$ is true, that means that $\dfrac 12 + \dfrac 23 + ... +\dfrac n{n+1} + \dfrac {n+1}{n+2} \lt \dfrac {n^2}{n+1} + \dfrac {n+1}{n+2}$

So if I can prove if $\dfrac {n^2}{n+1} + \dfrac {n + 1}{n+2} \ge \dfrac {\;(n+1)^2}{n+2}$, then by transitivity $P(n+1)$ is true.

But I get $3n\le 2n$ which isn't true

Answer 1

Hint : The largest summand of the left side is $\frac{n}{n+1}$. You do not need induction.

If you need to do induction, follow the hint given above.