91影视

The given series is$\underset{k=0}{\overset{鈭�}{鈭�}}\frac{1}{k^3+1}.$

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

So,

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

Let's solve denominator first,

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

Now,

$$\begin{gathered} 蚁=\frac{\underset{k鈫�鈭�}{\lim }1}{\underset{k鈫�鈭�}{\lim }\left({\left(k^3+1\right)}^{\frac{1}{k}}\right)} \\ 蚁=1 \end{gathered}$$

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 $\underset{k=1}{\overset{鈭�}{鈭�}}{a}_k\mathrm{and}\underset{k=1}{\overset{鈭�}{鈭�}}{\mathrm{b}}_k$be two series with positive terms such that $0鈮�a_k鈮�b_k$ for every positive number k.If the series $\underset{k=1}{\overset{鈭�}{鈭�}}{\mathrm{b}}_k$then the series $\underset{k=1}{\overset{鈭�}{鈭�}}{a}_k$also converges.

Now, let the series $\underset{k=0}{\overset{鈭�}{鈭�}}{a}_k=\underset{k=0}{\overset{鈭�}{鈭�}}\frac{1}{k^3+1}\mathrm{and}{\displaystyle \underset{k=0}{\overset{鈭�}{鈭�}}}{\mathrm{b}}_k=\underset{k=0}{\overset{鈭�}{鈭�}}\frac{1}{k^3}.$

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

As we can see $\underset{k=0}{\overset{鈭�}{鈭�}}{\mathrm{b}}_k=\underset{k=0}{\overset{鈭�}{鈭�}}\frac{1}{k^3}$is of the form $\underset{k=0}{\overset{鈭�}{鈭�}}{\mathrm{b}}_k=\underset{k=0}{\overset{鈭�}{鈭�}}\frac{1}{k^p}.$

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 $\underset{k=0}{\overset{鈭�}{鈭�}}{\mathrm{b}}_k=\underset{k=0}{\overset{鈭�}{鈭�}}\frac{1}{k^3}\mathrm{converges}\mathrm{than}\underset{k=0}{\overset{鈭�}{鈭�}}{a}_k=\underset{k=0}{\overset{鈭�}{鈭�}}\frac{1}{k^3+1}\mathrm{also}\mathrm{converges}.$

Hence, the given series converges.