Prove that $1<\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3n+1}$ Prove that $1<\dfrac{1}{n+1}+\dfrac{1}{n+2}+...+\dfrac{1}{3n+1}$.
By using the Mathematical induction. Suppose the statement holds for $n=k$.
Then for $n=k+1$. We have $\dfrac{1}{k+2}+\dfrac{1}{k+3}+...+\dfrac{1}{3k+1}+\dfrac{1}{3k+2}+\dfrac{1}{3k+3}+\dfrac{1}{3k+4}=(\dfrac{1}{k+1}+\dfrac{1}{k+2}+\dfrac{1}{k+3}+...+\dfrac{1}{3k+1})+(\dfrac{1}{3k+2}+\dfrac{1}{3k+3}+\dfrac{1}{3k+4}-\dfrac{1}{k+1})$
we know $\dfrac{1}{k+1}+\dfrac{1}{k+2}+\dfrac{1}{k+3}+...+\dfrac{1}{3k+1}>1$
What can we do for $(\dfrac{1}{3k+2}+\dfrac{1}{3k+3}+\dfrac{1}{3k+4}-\dfrac{1}{k+1})$?
 A: By cauchy-schwarz inequality
$$\left(\dfrac{1}{n+1}+\dfrac{1}{n+2}+\cdots+\dfrac{1}{3n+1}\right)(n+1+n+2+\cdots+3n+1)>(1+1+\cdots+1)^2=(2n+1)^2$$
note$$ (n+1+n+2+\cdots+3n+1)=\dfrac{(n+1+3n+1)(2n+1)}{2}=(2n+1)^2$$
A: You are almost done. Prove that 
$$\frac{1}{3k+2}+\frac{1}{3k+3}+\frac{1}{3k+4}-\frac{1}{k+1}$$ is positive.
To do this it is enough to show that $\frac{1}{3k+2}+\frac{1}{3k+4} \gt \frac{2}{3k+3}$. The left side can be written as $\frac{6k+6}{(3k+2)(3k+4}$. So we want to show that $(3k+3)^2\gt (3k+2)(3k+4)$. 
A: Another way: You can bound it by $$\frac{1}{n+1} + \frac{1}{n+2} + ... + \frac{1}{3n+1} \ge \int_{n+1}^{3n+1} \frac{1}{x} = \log \Big(\frac{3n+1}{n+1}\Big)$$ where $\frac{3n+1}{n+1}$ is monotone increasing in $n$, and $\log(\frac{3*7 + 1}{7+1}) > 1$ already. Then check the cases $n=1$ through $6$.
A: using $maxima$ I get

(%i1) 1/(3*k+2)+1/(3*k+3)+1/(3*k+4)-1/(k+1),ratexpand;
                                       2
(%o1)                      -------------------------
                               3       2
                           27 k  + 81 k  + 78 k + 24
(%i2) solve(denom(%),[k]);
                                 4                 2
(%o2)                     [k = - -, k = - 1, k = - -]
                                 3                 3

For $k=0$ the denominator is greater than $0$, so it is greater than $0$ for all $ k \gt -\frac{2}{3}$.
