Find $$ \sum_{n = 1}^{\infty} \frac{6 - 2^{2n - 1}}{3^n} $$

There are many ways to find that the limit is divergent, but the question explicitly states the sum must be interpreted as a geometric series. I know how to render the exponential terms to $(n-1)$; it's just the $(6 - \ldots)$ is bugging me -- I can't for the life of me find a way find to isolate the term so I get $a \cdot r^{n-1}$.

  • $\begingroup$ You can split it apart into two geometric series. As you did not quote the question I don't know if that is allowed. $\endgroup$ – Ross Millikan Feb 13 at 20:21

Expanding of Ross Millikan's comment, you could try \begin{align} \sum_{n = 1}^{\infty} \frac{6 - 2^{2n - 1}}{3^n} & = \sum_{n = 1}^{\infty} \frac{6}{3^n} - \sum_{n = 1}^{\infty} \frac{2^{2n - 1}}{3^n} = 2 \cdot \sum_{n = 1}^{\infty} \left( \frac{1}{3} \right)^{n - 1} - \frac{1}{2} \sum_{n = 1}^{\infty} \left( \frac{4}{3} \right)^n \\ & = 2 \cdot \sum_{n = 0}^{\infty} \left( \frac{1}{3} \right)^{n} - \frac{1}{2} \sum_{n = 1}^{\infty} \left( \frac{4}{3} \right)^n = 2 \cdot \frac{3}{2} - \frac{1}{2} \left( \sum_{n = 0}^{\infty} \left(\frac{4}{3} \right)^n - 1\right) \\ & = 3 + \frac{1}{2} - \frac{1}{2} \sum_{n = 0}^{\infty} \left( \frac{4}{3} \right)^n = \frac{7}{2} - \frac{1}{2} \sum_{n = 0}^{\infty} \left( \frac{4}{3} \right)^n = - \infty. \end{align}


A geometric series is a sires of the form: $$\sum_{n = 0}^{\infty} ar^n$$

And the only time such a series converges is when $|r|<1$. In that case, what it converges to can be found using the following formula:

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

To convert the given series into a geometric one (or ones), just start by shifting the index $n$ down by $1$ to make it begin at $0$:

$$ \sum_{n = 1}^{\infty} \frac{6 - 2^{2n - 1}}{3^n}= \sum_{n = 0}^{\infty} \frac{6 - 2^{2(n+1) - 1}}{3^n}= \sum_{n = 0}^{\infty} \frac{6 - 2^{2n+1}}{3^n}=\\ \sum_{n = 0}^{\infty} \left(\frac{6}{3^n}-\frac{2\cdot2^{2n}}{3^n}\right)= 6\sum_{n = 0}^{\infty} \left(\frac{1}{3}\right)^n-2\sum_{n = 0}^{\infty}\left(\frac{4}{3}\right)^n $$

The first of the two geometric series converges because $|1/3|<1$. The second geometric series diverges because $|4/3|$ is not less then $1$. If one series in a sum diverges, the whole thing diverges.


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