Q.2C) - Calculus

Question

A  p series other than $\sum _{k=1}^{\infty } \frac{1}{k^2}$you could use with comparison test to show that the series $\sum _{k=1}^{\infty }\frac{k-1}{k^3+k+1}$ converges.

Step-by-Step Solution

Answer

It is convergent

Step 1

Consider the series $\sum _{k=1}^{\infty }\frac{k-1}{k^3+k+1}$

To determine p series that used to show that $\sum _{k=1}^{\infty }\frac{k-1}{k^3+k+1}$ is convergent

The terms of series $\sum _{k=1}^{\infty }\frac{k-1}{k^3+k+1}$ are positive.

The series $\sum _{k=1}^{\infty }{b}_k$ for the series $\sum _{k=1}^{\infty }\frac{k-1}{k^3+k+1}$is

$\sum _{k=1}^{\infty }{b}_k=\sum _{k=1}^{\infty }\frac{1}{k^{3/2}}$

c) Step 2

The ratio $\underset{k\to \infty }{\lim }\frac{a_k}{b_k}$is given

$$\begin{gathered} \underset{k\to \infty }{\lim }\frac{a_k}{b_k}=\underset{k\to \infty }{\lim }\frac{{\displaystyle \frac{k-1}{k^3+k+1}}}{{\displaystyle \frac{1}{k^{3/2}}}} \\ =\underset{k\to \infty }{\lim }\frac{k^{3/2}(k-1)}{k^3+k+1} \\ =\underset{k\to \infty }{\lim }\frac{k^{5/2}(1+{\displaystyle \frac{1}{k}})}{k^3(1+{\displaystyle \frac{1}{k^2}}+{\displaystyle \frac{1}{k^3}})} \\ =\underset{k\to \infty }{\lim }\frac{(1+{\displaystyle \frac{1}{k}})}{k^{1/2}(1+{\displaystyle \frac{1}{k^2}}+{\displaystyle \frac{1}{k^3}})} \\ =0 \end{gathered}$$

c) Step 3

The value 0f $\underset{k\to \infty }{\lim }\frac{a_k}{b_k}=0$

The series $\sum _{k=1}^{\infty }{b}_k=\sum _{k=1}^{\infty }\frac{1}{k^{3/2}}$ is convergent by the p-series test

Then $\sum _{k=1}^{\infty }{a}_k$ is also convergent

Then the series $\sum _{k=1}^{\infty }\frac{k-1}{k^3+k+1}$is convergent and the p series is $\sum _{k=1}^{\infty }{b}_k=\sum _{k=1}^{\infty }\frac{1}{k^{3/2}}$