Solving for $\frac 13 +\frac 29 +\frac 3{27}+\cdots$

Question

Evaluate the sum $\frac{1}{3^1} + \frac{2}{3^2} + \frac{3}{3^3} + \cdots + \frac{k}{3^k} + \cdots $

How would I go about solving this problem? I'm thinking of setting the sum to $S$, multiplying by $3$, and then subtracting from the original equation. HELP!

I have explained my reasoning in the comment below.

Answer 1

So as you said, let's set

$$S=\frac{1}{3^1} + \frac{2}{3^2} + \frac{3}{3^3} + \cdots + \frac{k}{3^k}\cdots$$

Now we see

$$3S=\frac{1}{3^0} + \frac{2}{3^1} + \frac{3}{3^2} + \cdots + \frac{k}{3^{k-1}}\cdots$$

And so

$$3S-S=\frac{1-0}{3^0} + \frac{2-1}{3^1} + \frac{3-2}{3^2} + \cdots + \frac{k-(k-1)}{3^{k-1}}\cdots$$ or, rewritten,

$$2S=\frac1{3^0}+\frac1{3^1}+\frac1{3^2}+\cdots=\frac32$$

So that we get $S=\frac 34$.

Answer 2

Nice approach. You can also do this as follows and this method is useful for other types of series too. Compute the following sum $$f(x)=\sum_{k\geq 1}\left(\frac{x}{3}\right)^k$$ And then compute the derivative of $f$ and evaluate $f'(1)$.

Answer 3

Note that $$ \sum_{k=1}^\infty \frac{k}{3^k}=\sum_{k=1}^\infty \sum_{j=1}^k\frac{1}{3^k} =\sum_{j=1}^\infty \sum_{k=j}^\infty\frac{1}{3^k} =\sum_{j=1}^\infty\frac{1/3^j}{2/3} =\frac{3}{2}\frac{1/3}{1-(1/3)} =\frac{3}{4} $$ where the interchanging of summation is allowed since we are dealing with non-negative series.