Answer
The series converges by limit comparison.
Work Step by Step
$\Sigma_{n=1}^{\infty}\frac{10n+3}{n2^{n}}$
We can compare this series to the series $\Sigma_{n=1}^{\infty}\frac{10n}{n2^{n}}=10\Sigma_{n=1}^{\infty}\frac{1}{2^{n}}=10\Sigma_{n=1}^{\infty}(\frac{1}{2})^{n}$
The series $10\Sigma_{n=1}^{\infty}(\frac{1}{2})^{n}$ converges by the geometric series test, where $|r|=\frac{1}{2}$ and $0\lt\frac{1}{2}\lt1$.
If $\frac{10n+3}{n2^{n}}\leq(\frac{1}{2})^{n}$, then the original series also converges. This does not work, so instead let's use limit comparison:
$\lim\limits_{n \to \infty}\frac{a_n}{b_n}=L$
If $L$ is a finite and positive number, then either both series converge or both diverge, based on the convergence or divergence of $b_n$:
$\lim\limits_{n\to \infty}\frac{10n+3}{n2^{n}}\times\frac{2^n}{1}$
$\lim\limits_{n\to \infty}\frac{10n+3}{n}=\frac{\infty}{\infty}$
$\frac{\infty}{\infty}$ is an indeterminate form, so let's use L'hopital's rule.
$\lim\limits_{n \to \infty}\frac{10}{1}=10$
Since $10$ is a finite and positive integer, the original series converges because $10\Sigma_{n=1}^{\infty}(\frac{1}{2})^{n}$ converges.
Therefore, the original series converges by limit comparison.
