How to calculate $\sum_{n = 2}^{+\infty} \frac{5 + 2^n}{3^n}$? I need to calculate:
$$\sum_{n = 2}^{+\infty} \frac{5 + 2^n}{3^n}$$
Actually, I'm not particularly interested in knowing the result. What I'm really interested in is:


*

*What kind of series is it?

*How should I start calculating series like that?

 A: This is the sum of two geometric series:
$\displaystyle\sum_{n=2}^\infty \frac{5 + 2^n}{3^n} = 5\sum_{n=2}^\infty \left(\frac{1}{3}\right)^n + \sum_{n=2}^\infty \left(\frac{2}{3}\right)^n$.
A: Split it in two:
$$\sum_{n\ge 2}\frac{5+2^n}{3^n}=5\sum_{n\ge 2}\frac1{3^n}+\sum_{n\ge 2}\left(\frac23\right)^n$$
It’s a sum of two geometric series.
A: Your sum is the sum of two geometric series:
$$\sum_{n=2}^\infty \frac{5 + 2^n}{3^n} = 5\sum_{n=2}^\infty \left(\frac{1}{3}\right)^n + \sum_{n=2}^\infty \left(\frac{2}{3}\right)^n$$
In general, $$\sum_{n=0}^\infty ar^n = \frac{a}{1-r}$$ which converges when the magnitude of $r < 1$.
In your sum of two geometric series, for the first, we have $a = 5$, $r = \dfrac 13$,
so we start first with $$5\sum_{n=0}^\infty \left(\frac{1}{3}\right)^n = \frac{5}{1-1/3} = \frac{15}{2}$$
But your sum starts from $n=2$, not $n = 0$, so we start from the basic formula for evaluating a convergent geometric sum, and then subtract from that the value of the terms at $n = 0, n=1$ $$5\sum_{n=2}^\infty \left(\frac{1}{3}\right)^n\,.$$ 
Now we need to subtract from $\dfrac{15}{2}$ the evaluation of the terms when $n = 0, n = 1$:
$n = 0 \to \dfrac{5}{3^0} = 5$
$n = 1 \to \dfrac{5}{3}$
Hence, the sum of the first of the split sums evaluates to $\dfrac{15}{2} - 5 - \dfrac{5}{3} = \dfrac{5}{2} - \dfrac{5}{3} = \dfrac{5}{6}$
Similarly, you can compute the sum of the second of the split sums, using this strategy.
