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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

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    You got inequality reversed, you want $\frac{n^2}{n+1}+\frac{n+1}{n+2}<\frac{(n+1)^2}{n+2}$, not the other way around – 2017-02-28
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    Welcome to math stack exchange! – 2017-02-28
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    Just a question to the kind editor, what is the benefit of a dfrac over a frac in the typesetting? – 2017-02-28
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    @AnotherJohnDoe Using `\dfrac` instead of `\frac` in titles is almost always a bad idea. *Adding* it to a title is a quite bad idea. – 2017-02-28
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    Well, when I initially joined the site and tried using the \frac command, it didn't seem to work. A person then edited my post using \dfrac, and I've been using it since. – 2017-02-28
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    @AnotherJohnDoe "it didn't seem to work" Huh? Anyway... just don't. – 2017-02-28

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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.