Calculate exact value of and infinite sum

Question

Im trying to find the exact value of the infinite sum :

3 + 1/3 + 1/27 + 1/243 + 1/2187 + ...

I can see that to generate new terms we take the previous term and divide by 9 or multiply by 9. Not exactly sure if one way or the other makes it any easier. This is what i have come up with to solve the series is this correct and where do i go from here?

$\sum_{k=3}^\infty= K_0 + K_n(1/9)\\ $

Answer 1

Using the idea presented by @MarkBennet in Closed formula for the sum of the following series we first let:$$S=3+1/3+1/27+1/243+1/2187+...$$Then divide both sides by 9 to get:$$S/9=1/3+1/27+1/243+1/2187+1/19683+...=S-3$$Therefore:$$3=S-S/9=8S/9$$$$8S/9=3$$$$S=(9*3)/8=27/8$$

NOTE: This method only works for a series where the magnitude of the common ratio between terms is less than 1.

Answer 2

You are always dividing by $9$ in the examples given. Do you know how to sum a geometric series?You can say that if $S$ is the sum, $S=3+S/9$, as the divide by $9$ changes each term to the next.

Answer 3

By term-wise cancellation, $$\begin{align} S&=a+ab+ab^2+ab^3+ab^4...\\ Sb&=\ \ \ \ \ \ \ ab+ab^2+ab^3+ab^4...\\ S-Sb&=a \end{align}$$ we have$$S=\frac a{1-b}.$$ This is only valid when the sum converges, which requires $|b|<1$.

Applied to the given case, we have $$S=\frac3{1-\frac19}=\frac{27}8.$$