<s>[INST] For any $n \in \mathbb{N}$, show that: $\frac{1}{n+1} + \frac{1}{n+2} + \ldots + \frac{1}{2n} < \frac{5}{6}$. For any $n \in \mathbb{N}$, show that:
$$\frac{1}{n+1} + \frac{1}{n+2} + \ldots + \frac{1}{2n} < \frac{5}{6}$$
I wrote the sum as $H_{2n} - H_{n}$, where $H_{k} = \frac{1}{1} + \frac{1}{2} + \ldots + \frac{1}{k}$ (the kth harmonic number). After that, I was searching for inequalities with harmonic numbers, but I didn't find anything useful.
Can you, please, give me a hint? I don't want the full proof. Thank you!
 [/INST] We need to prove that
$$\frac{1}{n+1}-\frac{1}{n}+\frac{1}{n+2}-\frac{1}{n}+...+\frac{1}{n+n}-\frac{1}{n}<\frac{5}{6}-1$$ or
$$\frac{1}{n+1}+\frac{2}{n+2}+...+\frac{n}{n+n}>\frac{n}{6}.$$
Now, by C-S
$$\frac{1}{n+1}+\frac{2}{n+2}+...+\frac{n}{n+n}=$$
$$=\frac{1^2}{1(n+1)}+\frac{2^2}{2(n+2)}+...+\frac{n^2}{n(n+n)}\geq$$
$$\geq\frac{(1+2+...+n)^2}{1(n+1)+2(n+2)+...+n(n+n)}=$$
$$=\frac{\frac{n^2(n+1)^2}{4}}{n(1+2+...+n)+1^2+2^2+...+n^2}=$$
$$=\frac{\frac{n^2(n+1)^2}{4}}{\frac{n^2(n+1)}{2}+\frac{n(n+1)(2n+1)}{6}}=\frac{3n(n+1)}{2(5n+1)}>\frac{n}{6}.$$
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