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 $(\dfrac{1}{3k+2}+\dfrac{1}{3k+3}+\dfrac{1}{3k+4}-\dfrac{1}{k+1})$?

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}$.
