Question

I found the interval of convergence, but I am not sure how to do the second part, finding the sum of the series as a function of x.

Answer 1

The given series is geometric with common ratio $6^x - 9$, which converges if $|6^x - 9|<1$ (i.e. the interval of convergence). We have the well-known result

$\displaystyle |r| < 1 \implies \sum_{n=0}^\infty ar^n = \frac{a}{1-r}$

If you're not familiar with that result, it's easy to reproduce.

Let $S_N$ be the $N$-th partial sum of the infinite series,

$\displaystyle S_N = \sum_{n=0}^N \left(6^x - 9\right)^n = 1 + \left(6^x - 9\right) + \left(6^x - 9\right)^2 + \cdots + \left(6^x - 9\right)^N$

Multiply both sides by the ratio.

$\left(6^x - 9\right) S_N = \left(6^x - 9\right) + \left(6^x - 9\right)^2 + \left(6^x - 9\right)^3 + \cdots + \left(6^x - 9\right)^{N+1}$

Subtract this from $S_N$ to eliminate all the powers of the ratio between 0 and $N+1$.

$\left(1 - \left(6^x - 9\right)\right) S_N = 1 - \left(6^x - 9\right)^{N+1}$

Solve for $S_N$.

$S_N = \dfrac{1 - \left(6^x - 9\right)^{N+1}}{10-6^x}$

Now as $N\to\infty$, the exponential term converges to 0 and we're left with

$\displaystyle \sum_{n=0}^\infty \left(6^x-9\right)^n = \lim_{N\to\infty} S_N = \boxed{\frac1{10-6^x}}$