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Conditional Convergence – Definition, Condition, and Examples
Conditional convergence is an important concept that we need to understand when studying alternating series. Â If you want to master numerical analysis and fully understand series and sequence, it is essential that you know what makes conditionally convergent series unique.
Series that exhibit conditional convergence highlight the fact that it is possible for the original series to be convergent but the series form by its termsâ€™ absolute values may not be convergent.
In the past, weâ€™ve studied series with absolute convergence. Now, itâ€™s time that we learn about the condition we call the conditional series. These two concepts are closely related, so it will be helpful if you have your notes on absolute convergence handy.
Â Our discussion will cover the fundamentals of conditional convergence â€“ from its definition to understanding the process of checking whether a given series exhibits conditional convergence.
What is conditional convergence?
Conditional convergence tells us that it is possible for a series to be convergent but the series containing their termsâ€™ absolute values are divergent. Suppose that we have $\sum_{n = 0}^{\infty} a_n$, and we take the absolute value of each of the term. The resulting series will be as follows:
\begin{aligned}\sum_{n=0}^{\infty} a_n &= a_1 + a_2 + a_3 + â€¦\\\sum_{n = 0}^{\infty} |a_n| &= |a_1| + |a_2| + |a_3| + â€¦\end{aligned}
If the series, $\boldsymbol{\sum_{n = 0}^{\infty} a_n}$, is convergent, Â $\boldsymbol{\sum_{n = 0}^{\infty} |a_n|}$ is divergent, the series, $\boldsymbol{\sum_{n = 0}^{\infty} a_n}$ will exhibit conditional convergence.
Â A great example of a conditionally convergent series is the alternating harmonic series, $\sum_{n =1}^{\infty} (-1)^{n -1} \dfrac{1}{n}$.
\begin{aligned}\sum_{n =1}^{\infty} (-1)^{n -1} \dfrac{1}{n} &= 1- \dfrac{1}{2} + \dfrac{1}{3} – \dfrac{1}{4} + â€¦\end{aligned}
Sinc$\lim_{n \rightarrow \infty}(-1)^{n -1} \dfrac{1}{n} = 0$, through the alternating series test, we can confirm that this series is convergent. Now, letâ€™s take a look at the series containing the absolute values of this seriesâ€™ terms.
\begin{aligned}\sum_{n =1}^{\infty} \left|(-1)^{n -1} \dfrac{1}{n}\right| &= 1- \dfrac{1}{2} + \dfrac{1}{3} – \dfrac{1}{4} + â€¦\end{aligned}
As we have learned in the past, the resulting harmonic series is actually divergent. When this happens, the series, $\boldsymbol{\sum_{n =1}^{\infty} (-1)^{n -1} \dfrac{1}{n}}$ is conditionally convergent.
Conditional convergence test
Now, letâ€™s generalize this and set the conditions for the conditional convergence test. Suppose that we have a series, $\sum_{n = 0}^{\infty} a_n$, we can take the absolute values of its terms to see whether the series is conditionally convergent.
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â CONDITIONAL CONVERGENCE TEST When $\sum_{n = 0}^{\infty} a_n$ is convergent, but the series, $\sum_{n = 0}^{\infty} |a_n|$ is divergent (or not convergent), then the series, $\sum_{n = 0}^{\infty} a_n$ is conditionally convergent. |
Of course, as we have learned in the past when the resulting series is convergent, the original series is absolutely convergent. Â Keep these conditions in mind when working on the sample problems weâ€™ve prepared for you!
How to test for conditional convergence?
Testing a series for conditional convergence is actually similar to testing the series of absolute convergence. The only difference is our conclusion from the results. Â
- Confirm that our given series is convergent.
- Now, take the absolute values of each of the term of the series.
- See if the resulting series is convergent or not. Use the guide below for your conclusion:
\begin{aligned}\boldsymbol{\sum_{n =1}^{\infty} a_n }\end{aligned} | \begin{aligned}\boldsymbol{\sum_{n =1}^{\infty} |a_n| }\end{aligned} | Conclusion |
Convergent | Divergent | Conditionally Convergent |
Convergent | Convergent | Absolutely Convergent |
There are different tests we can apply when testing both series for convergence. Head over to this article for some of the common tests. There are instances when weâ€™ll need either the ratio or the root test as well, so check the links weâ€™ve provided in case you need a quick refresher. Now, test your understanding of this topic by answering the problems weâ€™ve provided below!
Example 1
Is the series, $\sum_{n =1}^{\infty} (-1)^n \dfrac{2n + 5}{3n^{2} + 4n + 5}$, is conditionally convergent, absolutely convergent, or divergent?
Solution
The series, $\sum_{n =1}^{\infty} (-1)^n \dfrac{2n + 5}{3n^{2} + 4n + 5}$, is an alternating series with terms that are decreasing as $n\rightarrow \infty$, so we can use the alternating series test to see if the series is convergent.
\begin{aligned}\lim_{n \rightarrow \infty}\dfrac{2n + 5}{3n^{2} + 4n + 5} &= \lim_{n \rightarrow \infty}\dfrac{\dfrac{2}{n} + \dfrac{5}{n^2}}{3 + \dfrac{4}{n} + \dfrac{5}{n^2}}\\&= 0\\&\Rightarrow \boldsymbol{\sum_{n = 1}^{\infty} (-1)^n\dfrac{2n + 5}{3n^{2} + 4n + 5}} \textbf{ is convergent}\end{aligned}
Weâ€™ve shown that our original series is convergent, itâ€™s time for us to see if $\sum_{n =1}^{\infty} \left|(-1)^n \dfrac{2n + 5}{3n^{2} + 4n + 5}\right|$ converges or diverges.
Â
\begin{aligned}\sum_{n =1}^{\infty} \left|(-1)^n \dfrac{2n + 5}{3n^{2} + 4n + 5}\right| &= \sum_{n =1}^{\infty} \dfrac{2n + 5}{3n^{2} + 4n + 5}\end{aligned}
The divergent test wonâ€™t be able to determine whether the series is divergent or not, so weâ€™ll use the comparison test â€“ the best series to use would the harmonic series, $\sum_{n =1}^{\infty} \dfrac{1}{n}$.
Â
For all values of $n$ within $(1, \infty)$, Â Â we can show that $\dfrac{1}{n} \leq \dfrac{2n + 5}{3n^{2} + 4n + 5}$.
\begin{aligned}\dfrac{1}{n} &\leq \dfrac{2n + 5}{3n^{2} + 4n + 5}\\ 3n^2 +4n + 5 &\leq 2n^2 + 5n\\5 &\leq n^2 + n\end{aligned}
Â We know that $\sum_{n =1}^{\infty} \dfrac{1}{n}$ is divergent, so through the comparison test, $\sum_{n =1}^{\infty} \dfrac{2n + 5}{3n^{2} + 4n + 5}$ is also divergent.
\begin{aligned}\boldsymbol{\sum_{n =1}^{\infty} (-1)^n \dfrac{2n + 5}{3n^{2} + 4n + 5}}\end{aligned} | \begin{aligned}\boldsymbol{\sum_{n =1}^{\infty} \left|(-1)^n \dfrac{2n + 5}{3n^{2} + 4n + 5}\right|} \end{aligned} | Conclusion |
Convergent | Divergent | Conditionally Convergent |
Hence, our series is conditionally convergent.
Â
Example 2
Is the series, $\sum_{n = 0}^{\infty} \dfrac{2^n}{5^n + 2^n}$, is conditionally convergent, absolutely convergent, or divergent?
Solution
Letâ€™s first take the limit of the $n$the term as $n$ approaches infinity. This will help us determine whether the series is divergent or not.
\begin{aligned} \lim_{n \rightarrow \infty} \dfrac{2^n}{5^n + 2^n} & =\lim_{n \rightarrow \infty} \dfrac{\dfrac{2^n}{2^n}}{\dfrac{5^n }{2^n} + \dfrac{2^n}{2^n}}\\&= \lim_{n \rightarrow \infty} \dfrac{1}{\left(\dfrac{5}{2} \right )^n + 1} \\&= \dfrac{1}{0 + 1}\\&= 1\\&\Rightarrow \boldsymbol{\sum_{n = 0}^{\infty} \dfrac{2^n}{5^n + 2^n}}\textbf{ is divergent} \end{aligned}
Through the divergent test, we can conclude that $\sum_{n = 0}^{\infty} \dfrac{2^n}{5^n + 2^n}$ is divergent.
This example is an important reminder to always double-check if the divergent test applies for the series (itâ€™s a straightforward test too, so it wonâ€™t be complicated!). If from this alone, weâ€™ve shown that the series is divergent, there is no need for us to check the series for absolute or conditional convergence.
Practice Questions
1. Is the series, $\sum_{n = 0}^{\infty} \dfrac{1}{3n^2 + 4n + 6}$, is conditionally convergent, absolutely convergent, or divergent?
2. Is the series, $\sum_{n = 1}^{\infty} (-1)^{n – 1} \dfrac{\ln n}{n}$ is conditionally convergent, absolutely convergent, or divergent?
3. Is the series, $\sum_{n = 1}^{\infty} (-1)^{n – 1} \dfrac{\ln n}{n}$ is conditionally convergent, absolutely convergent, or divergent?
4. Is the series, $\sum_{n = 1}^{\infty} (-1)^{n – 1} \dfrac{4n^2 + 1 }{3n^2 + 6n + 8}$ is conditionally convergent, absolutely convergent, or divergent?
Answer Key
1. $\sum_{n = 0}^{\infty} \dfrac{1}{3n^2 + 4n + 6}$ is absolutely convergent.
2. $\sum_{n = 1}^{\infty} (-1)^{n – 1} \dfrac{\ln n}{n}$ is conditionally convergent.
3. $\sum_{n = 1}^{\infty} (-1)^{n – 1} \dfrac{\tan^{-1}(n)}{n}$ is conditionally convergent.
4. $\sum_{n = 1}^{\infty} (-1)^{n – 1} \dfrac{4n^2 + 1 }{3n^2 + 6n + 8}$ is divergent