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. 

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
$$
1\lt\sum_{k=n}^{3n}\frac1{k+1}\lt2\tag8
$$
