Is the series $\sum_{n=1}^{\infty} \frac{4+3^n}{2^n}$ convergent or divergent $\sum_{n=1}^{\infty} \frac{4+3^n}{2^n}$
$\begin{align}
\sum_{n=1}^{\infty} \frac{4+3^n}{2^n} &= \sum_{n=1}^{\infty} \frac{4}{2^n} + \sum_{n=1}^{\infty} \frac{3^n}{2^n} \\
&= \sum_{n=1}^{\infty} \frac{4}{2 \cdot 2^{n-1}} + \sum_{n=1}^{\infty} \bigg(\frac{3}{2}\bigg)^n \\
&= 2\sum_{n=1}^{\infty}\frac{1}{2^{n-1}} + \sum_{n=1}^{\infty}\frac{3}{2} \bigg(\frac{3}{2}\bigg)^{n-1}
\end{align}$
Now I observe that the two terms are both geometric series and although the first one converges because $\frac{1}{2} < 1$, the second one doesn't because $\frac{3}{2} > 1$. Then can I assume that the series diverges and that is it?

Yes. Whenever you have a convergent series $\sum_{n=0}^\infty a_n$ and a divergent  series $\sum_{n=0}^\infty b_n$, the series $\sum_{n=0}^\infty(a_n+b_n)$ diverges.
Or you can apply the ratio test to reach the same conclusion.
