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.
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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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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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One Approach
Simpler than Cauchy-Schwarz is to note that for $x\in(n,n+1)$, $$ \frac1{n+1}\lt\frac1x\lt\frac1n\tag1 $$ which, when integrated in $x$ over $(n,n+1)$, gives $$ \frac1{n+1}\lt\log\left(\frac{n+1}n\right)\lt\frac1n\tag2 $$ then $$ \begin{align} \sum_{k=n}^{3n}\frac1{k+1} &\le\sum_{k=n}^{3n}\log\left(\frac{k+1}k\right)\\ &=\log\left(\frac{3n+1}n\right)\\[3pt] &=\log\left(3+\frac1n\right)\tag3 \end{align} $$ and $$ \begin{align} \sum_{k=n}^{3n}\frac1{k+1} &\ge\sum_{k=n}^{3n}\log\left(\frac{k+2}{k+1}\right)\\ &=\log\left(\frac{3n+2}{n+1}\right)\\[3pt] &=\log\left(3-\frac1{n+1}\right)\tag4 \end{align} $$ Since $\log(4)\lt2$, inequality $(3)$ shows the upper bound for $n\ge1$.
Since $\log\left(\frac{11}4\right)\gt1$, inequality $(4)$ shows the lower bound for $n\ge3$. We need only verify the lower bound for $n=1$ and $n=2$.
A Second Approach, Using Cauchy-Schwarz
Simply multiplying the number of terms by the largest term gives $$ \begin{align} \sum_{k=n}^{3n}\frac1{k+1} &\le\frac{2n+1}{n+1}\\ &=2-\frac1{n+1}\\[9pt] &\lt2\tag5 \end{align} $$ Applying Cauchy-Schwarz gives $$ \begin{align} \sum_{k=n}^{3n}\frac1{k+1}\sum_{k=n}^{3n}(k+1) &\gt\left(\sum_{k=n}^{3n}1\right)^2\\ &=(2n+1)^2\\ &=\sum_{k=n}^{3n}(k+1)\tag6 \end{align} $$ Equality does not hold in the first line of $(6)$ because the vectors are not parallel.
Therefore, $$ \sum_{k=n}^{3n}\frac1{k+1}\gt1\tag7 $$ Inequalities $(5)$ and $(7)$ show that $$ 1\lt\sum_{k=n}^{3n}\frac1{k+1}\lt2\tag8 $$
By C-S $$\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}+\frac{1}{2n+1}+\frac{1}{2n+2}+...+\frac{1}{3n}+\frac{1}{3n+1}=$$ $$=\left(\frac{1}{n+1}+\frac{1}{3n+1}\right)+\left(\frac{1}{n+2}+\frac{1}{3n}\right)+...+\left(\frac{1}{2n}+\frac{1}{2n+2}\right)+\frac{1}{2n+1}>$$ $$>\frac{(1+1)^2}{4n+2}+\frac{(1+1)^2}{4n+2}+...+\frac{(1+1)^2}{4n+2}+\frac{1}{2n+1}=1.$$ The right inequality we can prove by the similar way: $$\sum_{k=1}^{2n+1}\frac{1}{n+k}=\sum_{k=1}^n\left(\frac{1}{n+k}+\frac{1}{3n+2-k}\right)+\frac{1}{2n+1}=$$ $$=\sum_{k=1}^n\frac{4n+2}{(n+k)(3n+2-k)}+\frac{1}{2n+1}<\sum_{k=1}^n\frac{4}{3n}+\frac{1}{3}=\frac{5}{3}<2.$$