Sumation from 1 to infinity of [3^n + 4] / 2^n converges or diverges?
Given $\displaystyle \sum\limits_{n=1}^\infty \frac{3^n+4}{2^n}$, I guess you can apply direct comparison test here.
Let $\displaystyle b_n=\frac{3^n+4}{2^n}$, consider another sequence $\displaystyle a_n=\frac{3^n}{2^n}$. We have $\displaystyle 0\le a_n\le b_n,\;\;\forall n$. Since the geometric series $\displaystyle \sum\limits_{n=1}^\infty a_n=\sum\limits_{n=1}^\infty (\frac{3}{2})^n$ is obvious divergent. We can conclude that the given series $\displaystyle \sum\limits_{n=1}^\infty \frac{3^n+4}{2^n}$ must also be divergent.
$\displaystyle \sum_{n=1}^{\infty} \frac{3^n + 4}{2^n}$ Diverges, you can use the comparison test to show this, using $\displaystyle b_n \equiv \sum_{n=1}^{\infty} \frac{3^n}{2^n}=(\frac{3}{2})^n$ then using the root test on $\displaystyle b_n$, you get $\displaystyle \sqrt[n]{(\frac{3}{2})^n} = \frac{3}{2} > 1$, therefore it diverges