Question

In Exercises 29–34 use the ratio test to analyze whether the given series converges or diverges. If the ratio test is inconclusive, use a different test to analyze the series.

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

Short answer

The 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. Ratio Test.

Now,

$$\begin{gathered} \frac{a_{k+1}}{a_k}=\frac{\frac{1}{(k+1{)}^3+1}}{\frac{1}{k^3+1}} \\ =\frac{k^3+1}{(k+1{)}^3+1} \\ =\frac{k^3+1}{k^3+1+3k^2+3k+1} \\ \frac{a_{k+1}}{a_k}=\frac{k^3+1}{k^3+3k^2+3k+2} \\ \underset{k\to \infty }{\lim }\frac{a_{k+1}}{a_k}=\underset{k\to \infty }{\lim }\frac{k^3+1}{k^3+3k^2+3k+2} \\ =\left(\underset{k\to \infty }{\lim }\frac{k^3\left(1+\frac{1}{k^3}\right)}{k^3\left(1+\frac{3}{k}+\frac{3}{k^2}+\frac{2}{k^3}\right)}\right) \\ =1 \\ TestInconclusive. \end{gathered}$$

Step 3. Conclusion.

Now

$$\begin{gathered} \sum _{k=0}^{\infty }{b}_k=\sum _{k=0}^{\infty }\frac{1}{k^3}\text{is of the form}\sum _{k=0}^{\infty }{b}_k=\sum _{k=0}^{\infty }\frac{1}{k^p}\text{which is a}p\text{-series.} \\ \text{Here,}p=3>1\text{.} \end{gathered}$$

Hence, the series converges.