<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] Here's an alternative answer using $AM-HM$ inequality: $$\frac{a_1+a_2+\cdots+a_n}{n} \geq \frac{n}{\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}}$$
take $a_1=n+1, a_2=n+2,\cdots,a_{2n+1}=3n+1$, then by above inequality $$\frac{(n+1)+(n+2)+\cdots+(3n+1)}{2n+1} \geq \frac{2n+1}{\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{3n+1}}$$ Consequently, $$\frac{4n^2+4n+1}{2n+1} \geq \frac{2n+1}{\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{3n+1}}$$ which implies $$\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{3n+1} \geq \frac{(2n+1)^2}{4n^2+4n+1}=1$$ 
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