# 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?

• Don't even bother with all of that. Look at the limit of the summands... $\frac{4+3^n}{2^n}$, as $n$ gets large $4$ will get dwarfed in size by $3^n$, so this acts like $\frac{3^n}{2^n}$ which acts like $(\frac{3}{2})^n$ which grows without bound. Since the summands don't approach zero the series trivially diverges. – JMoravitz Sep 19 '19 at 22:16
• Yes, u can just use that it is a series of positive terms and $\frac{4+3^n}{2^n} > \frac{3^n}{2^n}$ – Dominik Kutek Sep 19 '19 at 22:16

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.
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 $$0$$.
• ... and that necessary condition is the sequence of the summands to converge to $0$. – peter.petrov Sep 19 '19 at 22:51