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Telescoping Series

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.

Telescoping Series

In this example, we will determine whether or not the series
\begin{align*}
\sum _{k =1}^{\infty}\frac{3}{3k-2}-\frac{3}{3k+1},
\end{align*}
converges or diverges. Furthermore, if the series does converge, we will find the sum.

In order to determine whether or not this converges, we will start with the first question in our decision tree,

  • Is \(\lim\limits_{k \to \infty}a_{k}=0\)?
    In this case, we see that
    \begin{align*}
    \lim_{k \to \infty}\frac{3}{3k-2}-\frac{3}{3k+1}=0-0=0,
    \end{align*}
    since the \(3k\) dominates the \(2\) (see Limits of Sequences for more on limits and dominating sequences).
  • Before continuing, I should note that we can indeed simplify the terms of the series. In particular, if we find a common denominator, we can turn these two fractions into one. If we do this, the answers to the following questions may be different, which means we do have another way to determine convergence here. However, we will instead leave the series as is and work through the decision tree.
  • 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?
    No, this does not look like a geometric series.
  • Does the terms look like a rational function?
    Referring to our earlier note, we could rewrite this to look like a rational function. However, as is, this is not a rational function.
  • Write a few terms. Do any of the terms cancel?
    We now are left with the task of writing out some terms and trying to determine what happens. In this case, we get that
    \begin{align*}
    \sum_{k=1}^{n}\frac{3}{3k-2}-\frac{3}{3k+1}&=\frac{3}{1}-\frac{3}{4}+\frac{3}{4}-\frac{3}{7}+\frac{3}{7}\\
    &-\ldots -\frac{3}{3n-2}+\frac{3}{3n-2}-\frac{3}{3n+1} \\
    &=3-\frac{3}{3n+1}.
    \end{align*}
    Therefore, many of the terms will cancel out.
  • Cancel and take the limit of the partial sums.
    We now find that
    \begin{align*}
    \sum_{k=1}^{\infty}\frac{3}{3k-2}-\frac{3}{3k+1}&=\lim_{n \to \infty}\sum_{k=1}^{n}\frac{3}{3k-2}-\frac{3}{3k+1} \\
    &=\lim_{n \to \infty}3-\frac{3}{3n+1} \\
    &=3-0=3.
    \end{align*}
    Hence, the series will converge to \(3\).

Conclusion

For this example, we saw that the terms did not look like a geometric sequence or a rational function. Therefore, our usual tests weren’t helpful. Instead, we had write out terms and cancel where appropriate so that we could take the limit of partial sums. Because this was a telescoping series, we were able to do just that.

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.

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