How to find the value of $\sum_{k=1}^\infty (\frac{1}{9})^k$ using partial sums?

Question

So I was trying to prove an infinite sum by looking at the partial sum, when I ran into a problem. Consider: $$\sum_{k=1}^n \left(\frac{1}{9}\right)^k = \frac{1}{8} 9^{-n} (9^n-1)$$

but as there are no solutions to $9^{-n} (9^n-1) = 1$, is it possible to prove that $$\sum_{k=1}^\infty \left(\frac{1}{9}\right)^k = \frac{1}{8}$$ by the partial sum?

Answer 1

You're right there: You want $\lim \limits_{n \to \infty} \frac{1}{8}(9^{-n} (9^n - 1))$

Distribute: $\lim \limits_{n \to \infty} \frac{1}{8}(1 - 9^{-n}) = \frac{1}{8}$ since $9^{-n} \to 0$ as $n \to \infty$

The problem you say that you "ran into" seems to be more of a conceptual issue. The problem I believe you're having is that there is no solution to $9^{-n}(9^n - 1) = 1$. This is true, but limits are not exact solutions always. Limits are what a sequence approaches. This means that we can get as close as we want, but that doesn't mean that we ever have to get there exactly. Thus, while there are no solutions to $9^{-n}(9^n - 1) = 1$, as $n \to \infty$ the left hand side approaches $1$, and that's good enough.

Answer 2

\begin{align*} \sum_{k = 1}^{\infty} \left(\frac{1}{9}\right)^k & = \lim_{n \to \infty} \sum_{k = 1}^{n} \left(\frac{1}{9}\right)^k\\ & = \lim_{n \to \infty} \frac{1}{8}9^{-n}(9^n - 1)\\ & = \frac{1}{8} \lim_{n \to \infty} (1 - 9^{-n})\\ & = \frac{1}{8}(1 - 0)\\ & = \frac{1}{8} \end{align*}