Question

Why can’t the comparison test make conclusions about ${\int }_2^{\infty }\frac{1}{{(x-1)}^2}$based on the convergence of${\int }_2^{\infty }\frac{1}{x^2}$? What graphical way could you argue for the convergence instead?

Short answer

The comparison test is not sufficient to make conclusions on the convergence or divergence of ${\int }_2^{\infty }\frac{1}{(x-1{)}^2}dx$based on the convergence of ${\int }_2^{\infty }\frac{1}{x^2}dx$because the relation between them cannot be defined.

Step-by-step solution

Step 1. Given information.   

We are given two integrals,

${\int }_2^{\infty }\frac{1}{{(x-1)}^2}$and

${\int }_2^{\infty }\frac{1}{x^2}$

Step 2. Comparing the integrals. 

Finding the relation between $\frac{1}{x^2}$and $\frac{1}{(x-1{)}^2}$in the interval $[2,\infty )$. $x^2<x^2+1$for all $x\in [2,\infty )$

$\frac{1}{x^2}>\frac{1}{x^2+1}$for all $x\in [2,\infty )$

Expand the term $\frac{1}{(x-1{)}^2}$by using $(a-b{)}^2=a^2+b^2-2ab$.

$\frac{1}{(x-1{)}^2}=\frac{1}{x^2+1-2x}$

$\frac{1}{x^2+1}<\frac{1}{x^2}$for all $x\in [2,\infty )$

Using comparison test, observe that ${\int }_2^{\infty }\frac{1}{x^2+1}dx$converges since ${\int }_2^{\infty }\frac{1}{x^2}dx$converges and $\frac{1}{x^2+1}<\frac{1}{x^2}$for all $x\in [2,\infty )$.

But, one cannot define the relation between $\frac{1}{x^2+1-2x}$and $\frac{1}{x^2}$for $x\in [2,\infty )$.