Question

In Exercises 35–40 use the root test to analyze whether the given series converges or diverges. If the root test is inconclusive, use a different test to analyze the series.

$\sum _{k=0}^{\infty }\frac{1}{k^3+1}$

Short answer

The given series converges.

Step-by-step solution

Step 1. Given Information.

The given series is$\sum _{k=0}^{\infty }\frac{1}{k^3+1}.$

Step 2. Determine whether the given series converges or diverges. 

By using the root test, let the general term is$a_k=\frac{1}{k^3+1}.$

So,

$$\begin{gathered} \rho =\underset{k\to \infty }{\lim }{\left(a_k\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$k$}\right.} \\ \rho =\underset{k\to \infty }{\lim }{\left(\frac{1}{k^3+1}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$k$}\right.} \\ \rho =\underset{k\to \infty }{\lim }\left(\frac{1}{{\left(k^3+1\right)}^{{\displaystyle \frac{1}{k}}}}\right) \\ \rho =\frac{\underset{k\to \infty }{\lim }1}{\underset{k\to \infty }{\lim }\left({\left(k^3+1\right)}^{\frac{1}{k}}\right)} \end{gathered}$$

Let's solve denominator first,

$$\begin{gathered} \underset{k\to \infty }{\lim }\left({\left(k^3+1\right)}^{\frac{1}{k}}\right) \\ \mathrm{Use}{a}^x=e^{\ln \left(a^x\right)} \\ \mathrm{So},{\left({\mathrm{k}}^3+1\right)}^{\frac{1}{\mathrm{k}}}={\mathrm{e}}^{\ln \left({\left({\mathrm{k}}^3+1\right)}^{\frac{1}{\mathrm{k}}}\right)} \\ {\left({\mathrm{k}}^3+1\right)}^{\frac{1}{\mathrm{k}}}={\mathrm{e}}^{\frac{1}{\mathrm{k}}\ln \left({\mathrm{k}}^3+1\right)} \\ \underset{\mathrm{k}\to \infty }{\lim }\left({\left({\mathrm{k}}^3+1\right)}^{\frac{1}{\mathrm{k}}}\right)=\underset{\mathrm{k}\to \infty }{\lim }\left({\mathrm{e}}^{\frac{1}{\mathrm{k}}\ln \left({\mathrm{k}}^3+1\right)}\right) \\ \underset{\mathrm{k}\to \infty }{\lim }\left({\left({\mathrm{k}}^3+1\right)}^{\frac{1}{\mathrm{k}}}\right)=\left({\mathrm{e}}^{\underset{\mathrm{k}\to \infty }{\lim }\frac{1}{\mathrm{k}}\ln \left({\mathrm{k}}^3+1\right)}\right) \\ \underset{\mathrm{k}\to \infty }{\lim }\left({\left({\mathrm{k}}^3+1\right)}^{\frac{1}{\mathrm{k}}}\right)={\mathrm{e}}^0 \\ \underset{\mathrm{k}\to \infty }{\lim }\left({\left({\mathrm{k}}^3+1\right)}^{\frac{1}{\mathrm{k}}}\right)=1 \end{gathered}$$

Now,

$$\begin{gathered} \rho =\frac{\underset{k\to \infty }{\lim }1}{\underset{k\to \infty }{\lim }\left({\left(k^3+1\right)}^{\frac{1}{k}}\right)} \\ \rho =1 \end{gathered}$$

Since it is1,thus the root test is inconclusive.

Step 3. Using the different test to analyze the series.

We will use the comparison test to analyze the series. The comparison test states that let $\sum _{k=1}^{\infty }{a}_k\mathrm{and}\sum _{k=1}^{\infty }{\mathrm{b}}_k$be two series with positive terms such that $0\le a_k\le b_k$ for every positive number k.If the series $\sum _{k=1}^{\infty }{\mathrm{b}}_k$then the series $\sum _{k=1}^{\infty }{a}_k$also converges.

Now, let the series $\sum _{k=0}^{\infty }{a}_k=\sum _{k=0}^{\infty }\frac{1}{k^3+1}\mathrm{and}{\displaystyle \sum _{k=0}^{\infty }}{\mathrm{b}}_k=\sum _{k=0}^{\infty }\frac{1}{k^3}.$

So, $0\le \frac{1}{k^3+1}\le \frac{1}{k^3}.$

As we can see $\sum _{k=0}^{\infty }{\mathrm{b}}_k=\sum _{k=0}^{\infty }\frac{1}{k^3}$is of the form $\sum _{k=0}^{\infty }{\mathrm{b}}_k=\sum _{k=0}^{\infty }\frac{1}{k^p}.$

The p-series say that if p > 1 then it converges and if p < 1 then it diverges.

Here, $p=3>1$thus, the series converges.

If $\sum _{k=0}^{\infty }{\mathrm{b}}_k=\sum _{k=0}^{\infty }\frac{1}{k^3}\mathrm{converges}\mathrm{than}\sum _{k=0}^{\infty }{a}_k=\sum _{k=0}^{\infty }\frac{1}{k^3+1}\mathrm{also}\mathrm{converges}.$

Hence, the given series converges.