Q. 31 - Calculus

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. 

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

Step-by-Step Solution

Answer

The series converges.

Step 1. Given information.

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

Step 2. Ratio Test.

Now,

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

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.