<s>[INST] Proving $\frac{1}{n+1} + \frac{1}{n+2}+\cdots+\frac{1}{2n} > \frac{13}{24}$ for $n>1,n\in\Bbb N$ by Induction Proving $\frac{1}{n+1} + \frac{1}{n+2}+\cdots+\frac{1}{2n} > \frac{13}{24}$ for $n>1,n\in\Bbb N$
To solve it I used induction but it is leading me nowhere my attempt was as follows:
Lets assume the inequality is true for $n = k$ then we need to prove that it is true for $k+1$
so we need to prove $\frac1{k+2} + \frac1{k+3}+\cdots+\frac1{2(k+1)} > 13/24$
I don't know where to go from here please help. 
 [/INST] $$f(n)=\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2n}=H_{2n}-H_n $$
is an increasing function, since
$$ f(n+1)-f(n) = \frac{1}{(2n+1)(2n+2)}>0. $$
It follows that $\forall n>1,\; f(n)\geq f(2)=\frac{7}{12}>\frac{13}{24}$.
Additionally, by Riemann sums
$$ \lim_{n\to +\infty}f(n)=\lim_{n\to +\infty}\sum_{k=1}^{n}\frac{1}{n+k} = \int_{0}^{1}\frac{dx}{1+x}=\log(2)$$
and
$$ 0< \int_{0}^{1}\frac{x^3(1-x)^3}{1+x}\,dx = \frac{111}{20}-8\log(2) $$
leads to $\log(2)<\frac{111}{160}$, so:
$$\boxed{ \forall n\geq 2,\quad \color{red}{\frac{7}{12}}\leq\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2n}<\color{red}{\frac{111}{160}}}$$
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