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.

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

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

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.

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.