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$\sum_{n=1}^{\infty} \frac{4+3^n}{2^n}$

$\begin{align} \sum_{n=1}^{\infty} \frac{4+3^n}{2^n} &= \sum_{n=1}^{\infty} \frac{4}{2^n} + \sum_{n=1}^{\infty} \frac{3^n}{2^n} \\ &= \sum_{n=1}^{\infty} \frac{4}{2 \cdot 2^{n-1}} + \sum_{n=1}^{\infty} \bigg(\frac{3}{2}\bigg)^n \\ &= 2\sum_{n=1}^{\infty}\frac{1}{2^{n-1}} + \sum_{n=1}^{\infty}\frac{3}{2} \bigg(\frac{3}{2}\bigg)^{n-1} \end{align}$

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?

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    $\begingroup$ 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. $\endgroup$ – JMoravitz Sep 19 '19 at 22:16
  • $\begingroup$ Yes, u can just use that it is a series of positive terms and $\frac{4+3^n}{2^n} > \frac{3^n}{2^n}$ $\endgroup$ – Dominik Kutek Sep 19 '19 at 22:16
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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.

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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 $0$.

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

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