1
$\begingroup$

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 $a \cdot r^{n-1}$.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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}

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.