Find sum of $\sum_{n=1}^{\infty}{\frac{2^{n+1}-8}{3^n}}$

Question

I want to find the sum of the following series:

$$\sum_{n=1}^{\infty}{\frac{2^{n+1}-8}{3^n}}$$

I have the intuition that this is a geometric series judging by the $n$ exponent in the numerator and denominator. Therefore I try to simplify a little bit:

$$\sum_{n=1}^{\infty}{\frac{2\cdot2^n-2^3}{3^n}}=\dots$$

I'm stuck here... I know that I need to group something but I'm not sure how. Preferably I want to end up with something like $\left(\frac{2}{3}\right)^n$.

Any hints on how to proceed to find the sum of the geometric series (which I know is equal to $\frac{1}{1-q}$ for $n=0$?

Answer 1

Hint:

First, use the fact that $$\frac{a+b}{c} = \frac ac + \frac bc$$

and then, use the fact that if $$\sum_{i=1}^\infty a_n\\\sum_{i=1}^\infty b_n$$ both converge, then $$\sum_{i=1}^\infty (a_n+b_n) = \sum_{i=1}^\infty a_n + \sum_{i=1}^\infty b_n$$

---

Additional facts to use:

• $\frac{a^n}{b^n} = \left(\frac ab\right)^n$
• $\sum_{i=1}^\infty (\alpha\cdot a_n) = \alpha\cdot \sum_{i=1}^\infty a_n$

Answer 2

Hint: $$\sum_{n=1}^{\infty}{\frac{2^{n+1}-8}{3^n}} = 2\sum_{n=1}^{\infty}{\frac{2^{n}}{3^n}} -8 \sum_{n=1}^{\infty}{\frac{1}{3^n}}$$ One can recognize the sum of a sequence of ratio $-1 < 2/3 < 1$ and another of ratio $-1 < 1/3 < 1$. (the interval $R<1$ guarantees the existence of a convergent solution)