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Divergence 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.

Divergent Series

In this example, we will determine whether or not the series
\begin{align*}
\sum _{k =1}^{\infty}\frac{2k^{2}+1}{\sqrt{k^{3}+2}},
\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 note that \(2k^{2} >> 1\) and \(k^{3} >> 2\) (see Limits of Sequences for more on limits and dominating sequences), therefore,
    \begin{align*}
    \lim_{k \to \infty}\frac{2k^{2}+1}{\sqrt{k^{3}+2}}&=\lim_{k \to \infty}\frac{2k^{2}}{\sqrt{k^{3}}} \\
    &=\lim_{k \to \infty}2k^{\frac{1}{2}} \\
    &=\infty.
    \end{align*}
    Therefore, the the terms become arbitrarily large and will not converge to 0.
  • The series diverges by the divergence test.

Conclusion

In this example, as long as we correctly found the limit of the terms, we were very quickly able to show that the series diverged. This is why, each time we are presented with a series, the first step is to check what happens to the terms. Because we did so, we should be able to move to the next exam problem without having to spend a lot of time here.

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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