<s>[INST] 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?
 [/INST] Your answer is correct!
You can also argue that $
\frac{4+3^{n}}{2^{n}} \underset{n \rightarrow \infty}{\longrightarrow} \infty
$ that is the necessary condition for a series to converge doesn't hold in our case that is- the limit of a sequence defining a convergent series is $0$. Or the limit of a general term in the series is $0$. Or as peter kindly commented-  the sequence of the summands to converge to  $0$.
</s>