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Root Test

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.

Root Test

In this example, we will determine whether or not the series
\begin{align*}
\sum _{k =2}^{\infty}\left(\frac{k}{2k+3}\right)^{2k},
\end{align*}
converges or diverges.

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}\left(\frac{k}{2k+3}\right)&=\frac{1}{2},
    \end{align*}
    therefore, our sequence will look like \(\frac{1}{2}^{2k}\) as \(k\) goes to \(\infty\). Therefore, these terms will tend to 0.
  • 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 the series look like a geometric series?
    In this case, it does look like we multiply by something in order to find each new term.
  • Can you rewrite the terms so that they look exactly like \(ar^{k}\)?
    Here we will not be able to do this.
  • Can you take the \(k\)th root of the terms?
    We see that
    \begin{align*}
    \sqrt[k]{\left(\frac{k}{2k+3}\right)^{2k}}&=\sqrt[k]{\left(\left(\frac{k}{2k+3}\right)^{2}\right)^{k}} \\
    &=\left(\frac{k}{2k+3}\right)^{2} \\
    &=\frac{k^{2}}{4k^{2}+12k+9}.
    \end{align*}
    Hence, we can take the \(k\)th root.
  • In order to answer the next question, we need to find that
    \begin{align*}
    \lim_{k \to \infty}\sqrt[k]{\left(\frac{k}{2k+3}\right)^{2k}}&=\lim_{k \to \infty}\frac{k^{2}}{4k^{2}+12k+9} \\
    &=\lim_{k \to \infty}\frac{k^{2}}{4k^{2}} \\
    ^=\frac{1}{4}.
    \end{align*}
    We now note that \(|\frac{1}{4}|< 1\), so the series will converge by the root test.

Conclusion

For this example, we saw that the terms looked like a geometric series. Because we were able to take the \(k\)th root of the terms, we were able to find show that the given series converged by using the root test.

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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