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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.

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  • $\begingroup$ Your idea seems good... What did it lead to? $\endgroup$ – vrugtehagel Mar 10 '18 at 22:59
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    $\begingroup$ Let the sum be $S$. This series looks almost geometric, but not quite. We can turn it into a geometric series as follows: \begin{align*} S &= \frac{1}{3^1} +\frac{2}{3^2} + \frac{3}{3^3} + \frac{4}{3^4} + \cdots \\ \frac{1}{3}S &= \frac{0}{3^1} + \frac{1}{3^2} + \frac{2}{3^3} + \frac{3}{3^4} + \cdots \\ \frac{2}{3}S = S - \frac{1}{3}S &= \frac{1}{3^1} + \frac{1}{3^2} + \frac{1}{3^3} + \frac{1}{3^4} + \cdots \end{align*}Now, we do have a geometric series, so we can find $\frac{2}{3}S = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{1}{2}$, and $S = \boxed{\frac{3}{4}}$. $\endgroup$ – A Piercing Arrow Mar 10 '18 at 23:02
  • $\begingroup$ That looks good! You basically had the answer already ;) $\endgroup$ – vrugtehagel Mar 10 '18 at 23:04
  • $\begingroup$ This is an arithmetico-geometric sequence: en.wikipedia.org/wiki/Arithmetico%E2%80%93geometric_sequence $\endgroup$ – an4s Mar 10 '18 at 23:19
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    $\begingroup$ Is it too much to ask you to write what you wrote in a comment, in the actual question you posted? $\endgroup$ – Namaste Mar 10 '18 at 23:32
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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 $S=\frac 34$.

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  • $\begingroup$ So I was correct! Thank you. $\endgroup$ – A Piercing Arrow Mar 10 '18 at 23:05
  • $\begingroup$ Yup, you were! Be sure to accept one of the answers so that this question doesn't end up in the unanswered section ;) $\endgroup$ – vrugtehagel Mar 10 '18 at 23:06
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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 $f'(1)$.

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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.

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