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.

**Alternating Series**In this example, we will determine whether or not the series

\begin{align*}

\sum _{k =2}^{\infty}\frac{(-1)^{k}}{k\ln(k)},

\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 even though some terms are positive and some terms negative, the bottom will dominate the top, thus giving us that the terms will tend to \(\pm 0\). Since \(0=-0\), all the terms tend to 0. - Is this an alternating series?

Note that there is a term of the form \((-1)^{k}\) in the terms of the series, so it is an alternating series. - The series converges by the alternating series test. Do you know if \(\sum |a_{k}|\) converges?

No, we have not determined this yet. - Start over to determine if the series \(\sum |a_{k}|\) converges or diverges.

**Integral Test**

We are now starting over to look at the series of the absolute value of the terms. If this converges, we will have an absolutely convergent series. However, if the series of absolute value of terms does not converge, then we will have a conditionally convergent series.

We, therefore, determine if

\begin{align*}

\sum _{k =2}^{\infty}\frac{1}{k \ln(k)},

\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 again see that the terms tend to 0. - Is this an alternating series?

Since we are focused on the \(|a_{k}|\), we do not have an alternating sereis. - Does the series look geometric?

It does not look like we are multiplying by something in order to determine the next term. - Do the terms look like a rational function?

While there is a \(\ln(k)\) in the terms, this does look nearly rational. - Do the terms of the form \(\frac{1}{k^{p}}\)?

No, these terms do not look exactly like this. - Are the terms exactly rational?

No, as previously stated, we have a \(\ln(k)\) in the terms. - Can you integrate the function?

In this situation, we will look at \(\int \frac{1}{x\ln(x)}dx\). In order to determine if we can integrate this, we will use our Techniques of Integration Decision Tree.- Can you guess the answer?

No, we cannot guess the answer here. - Is there an inside function?

Yes, we can use \(\ln(x)\) as a function inside of the function \(\frac{1}{i}\). - We then substitute \(i=\ln(x)\), \(di=\frac{1}{x}dx\) and \(dx=xdi\) in order to find,

\begin{align*}

\int \frac{1}{x\ln(x)}dx &=\int \frac{1}{xi}xdi \\

&=\int \frac{1}{i}di \\

&=\ln|i|+c \\

&=\ln|\ln(x)|+c.

\end{align*}

Hence, we can find the integral of the function.

- Can you guess the answer?
- We now have to use the integral test. We, therefore, find that

\begin{align*}

\int_{2}^{\infty}\frac{1}{x\ln(x)}dx&=\lim_{b \to \infty}\int_{2}^{b}\frac{1}{x\ln(x)}dx \\

&=\lim_{b \to \infty}\ln|\ln(x)| |_{2}^{b} \\

&=\lim_{b \to \infty}\ln(\ln(b))-\ln(\ln(2)) \\

&=\infty.

\end{align*}

Hence, the integral diverges, so the series diverges. - Since the series converges, but \(\sum |a_{k}|\) diverges, we find that the series is conditionally convergent.

**Conclusion**

In this example, we were very quickly able to show that the series would converge since it was an alternating series whose terms go to 0. However, we had to do more work in order to determine if it was absolutely or conditionally convergent. In that case, we had to use the integral test by finding an improper integral of a related function. We found that the absolute value of the series was divergent by the integral test. Therefore, the original series conditionally converged.

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.