Activity 8.3.9.
Consider the series defined by
\begin{equation}
\sum_{k=1}^{\infty} \frac{2^k}{3^k-k}\text{.}\tag{8.3.4}
\end{equation}
This series is not a geometric series, but this activity will illustrate how we might compare this series to a geometric one. Recall that a series \(\sum a_k\) is geometric if the ratio \(\frac{a_{k+1}}{a_k}\) is always the same. For the series in (8.3.4), note that \(a_k = \frac{2^k}{3^k-k}\text{.}\)
(a)
To see if \(\sum \frac{2^k}{3^k-k}\) is comparable to a geometric series, we analyze the ratios of successive terms in the series. Complete the table below, listing your calculations to at least 8 decimal places.
| \(k\) | \(\dfrac{a_{k+1}}{a_k}\) |
| \(5\) | |
| \(10\) | |
| \(20\) | |
| \(21\) | |
| \(22\) | |
| \(23\) | |
| \(24\) | |
| \(25\) |
(b)
Based on your calculations in the table in part 8.3.9.a, what can we say about the ratio \(\frac{a_{k+1}}{a_k}\) if \(k\) is large?
(c)
Do you agree or disagree with the statement: “the series \(\sum \frac{2^k}{3^k-k}\) is approximately geometric when \(k\) is large”? If not, why not? If so, do you think the series \(\sum \frac{2^k}{3^k-k}\) converges or diverges? Explain.
