How to prove $\sum_{n=1}^{\infty} \frac{3^n +7n}{2^n (n^2+1)} $ diverges? $$\sum_{n=1}^{\infty} \frac{3^n +7n}{2^n (n^2+1)} $$

It seems clear to me that this seires diverges since the dominant term is $(3/2)^n$, a geometric series with $r &gt; 1$

However I am required to prove this using convergence tests, presumably the comparison test to prove divergence.

I cannot work out a suitable comparison to make to prove divergence, suggestions?
you can prove that the term doesn't tend to 0, so the necessary condition of convergence is not satisfied, and since all terms are positive, the sum has to be $\infty$.

If you want to compare:
clearly $\frac{3^n +7n}{2^n (n^2+1)}&gt;\frac{3^n}{2^n (n^2+1)}\ge \frac{3^n }{2^n (n^2+n^2)}$ and now you can easily use Cauchy's test to prove that $\sum\frac{3^n }{2^n \cdot 2n^2}$ diverges.
Note that 

$$ \frac{3^n +7n}{2^n (n^2+1)} \ge \frac{3^n }{2^n (n^2+1)}, $$

However, $$\sum_{n=1}^{\infty}\frac{3^n}{2^n (n^2+1)} $$ diverges, so the original serries cannot converges.