Solved with jatloe12345 For $3n+1$ to be prime, $n$ must be even. Note that the number is equivalent to $$1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots + \frac{1}{2n-1} + \frac{1}{2n} - 2 ( \frac{1}{2} + \frac{1}{4} + \ldots + \frac{1}{2n})$$And this runs from $\frac{1}{n+1}$ to $\frac{1}{2n}$. Hence, we can pair off the terms as follows: $$(\frac{1}{n+1} + \frac{1}{2n}) + (\frac{1}{n+2} + \frac{1}{2n-1}) + \ldots + (\frac{1}{\frac{3}{2} n} + \frac{1}{\frac{3}{2}n + 1})$$$$= \frac{3n+1}{(2n)(n+1)} + \frac{3n+1}{(2n-1)(n+2)} + \ldots + \frac{(3n+1)}{(\frac{3}{2}n)(\frac{3}{2}n + 1)}$$Which is a multiple of $3n+1 \quad \blacksquare$