<s>[INST] Prove $\ 1<\frac{1}{n+1}+\frac{1}{n+2}+ ... + \frac{1}{3n+1}<2 $ using Cauchy-Schwarz How do I prove this inequality 
$$\ 1<\frac{1}{n+1}+\frac{1}{n+2}+ ... + \frac{1}{3n+1}<2 $$
I've tried to prove that $$\ \frac{1}{n+1}+\frac{1}{n+2}+ ... + \frac{1}{3n+1} \ $$ is less than $$\ \frac{1}{1}+\frac{1}{2}+ ... + \frac{1}{n} \ $$ and that this sum is either less than $2$ or equal to $2$ using C-S. 
 [/INST] Hint.
Define 
$$
f_n(x) = \sum_{k=1}^{2n+1}x^{n+k-1}=x^n\left(\frac{x^{2n+1}-1}{x-1}\right)
$$
hence
$$
\int f_n(x) dx = \int x^n\left(\frac{x^{2n+1}-1}{x-1}\right)dx = \frac{x^{n+1}}{n+1}+\cdots+\frac{x^{3n+1}}{3n+1}
$$
and now apply the Cauchy-Schwartz inequality on $u(x) = x^n, v(x) = \frac{x^{2n+1}-1}{x-1}$.
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