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

Question

$$\sum_{n=1}^{\infty} \frac{3^n +7n}{2^n (n^2+1)} $$

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

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?

Answer 1

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.

Answer 2

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 $\frac{3^n +7n}{2^n (n^2+1)}>\frac{3^n}{2^n (n^2+1)}\ge \frac{3^n }{2^n (n^2+n^2)}$ and now you can easily use Cauchy's test to prove that $\sum\frac{3^n }{2^n \cdot 2n^2}$ diverges.

Answer 3

You should consider the limit of the $n$th term as $n \to \infty$. As your intuition has directed you, the $(3/2)^n$ term dominates the limit, and the limit is infinite. [Alternately, two applications of l'Hopital's rule will work as well].

As the limit of the terms isn't $0$, the series cannot converge.