Through the following posts, we will be looking at different series. Our goal will be to determine if the series converges or diverges. Furthermore, if the series converges and we can find the actual value, we will. These posts are meant to help as you study series in Calculus 2. As you work through these problems with us, we hope you will make use of the Series Convergence Test Decision Tree that we have made available. You can find more series and other calculus study help posts at Calculus study help. Furthermore, there are videos meant to accompany many of these posts on YouTube.

**A Geometric Series**

For the our first example, we will try to determine if the series \(\sum\limits_{k=1}^{\infty}\frac{2^{k+2}}{3^{k}}\) converges or diverges. Before we start to work through our decision tree, I would take the time to simplify this and write it as

\begin{align*}

\sum\limits_{k=1}^{\infty}\frac{2^{k+2}}{3^{k}}&=\sum\limits_{k=1}^{\infty}4\frac{2^{k}}{3^{k}}.

\end{align*}

We now start with the first question,

- Is \(\lim\limits_{k \to \infty}a_{k}=0\)?

In this case, we see that

\begin{align*}

\lim_{k \to \infty}4 \frac{2^{k}}{3^{k}}=0,

\end{align*}

since the \(3^{k}\) dominates the \(2^{k}\) (see Limits of Sequences for more on limits and dominating sequences). - Is this an alternating series?

Note that there is no term of the form \((-1)^{k}\) in the terms of the series, so it is not an alternating series. - Does this look like a geometric series?

Yes, this does look like we are finding the new terms by multiplying the previous by terms by something. - Can you rewrite the terms so that they are exactly in the form \(ar^{k}\)?

Yes, through a little more simplification, we can arrive at \(4 (\frac{2}{3})^{k}\). - Is \(|r| < 1\)?

Yes \(\frac{2}{3} < 1\). - The series converges to \(\frac{a}{1-r}\) where \(a\) is the first term.
- We therefore get that this series will converge. Since the first term of the series is \(4\left(\frac{2}{3}\right)^{1}\), we will furthermore get that the series converges to \(\frac{\left(\frac{8}{3}\right)}{1-\frac{2}{3}}=8\).

**Conclusion**

We were able to work through our decision tree until we arrived at an answer. In the end our series converged. Furthermore, since this was a geometric series, we were able to determine that the series converged precisely to 8.

Make sure to continue working through the problems presented in the other posts so that you can work on more types of series. Also, if you find these helpful, please like the posts and share them with anyone else that may be studying series.