Comparison and Limit Comparison Tests — Question 6

Question 6 – Solution

Step 1: Bound the denominator.

Clearly $n^2-n+1\le n^2$. Also $$n^2-n+1\ge\frac{n^2}{2} \iff n^2-2n+2=(n-1)^2+1\ge0.$$ Thus $n^2/2\le n^2-n+1\le n^2$.

Step 2: Take reciprocals carefully.

All expressions are positive, so reciprocals reverse the inequalities: $$\frac1{n^2}\le\frac1{n^2-n+1}\le\frac2{n^2}.$$ The upper bound is the decisive one for convergence. Since $\sum2/n^2$ converges, Direct Comparison proves convergence.

Step 3: Confirm asymptotically.

$$\lim_{n\to\infty}\frac{1/(n^2-n+1)}{1/n^2} =\lim_{n\to\infty}\frac1{1-1/n+1/n^2}=1.$$