How to calculate $\sum_{n = 2}^{+\infty} \frac{5 + 2^n}{3^n}$?

Question

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?

Answer 1

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.

Answer 2

This is the sum of two geometric series:

$\displaystyle\sum_{n=2}^\infty \frac{5 + 2^n}{3^n} = 5\sum_{n=2}^\infty \left(\frac{1}{3}\right)^n + \sum_{n=2}^\infty \left(\frac{2}{3}\right)^n$.

Answer 3

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.