Comparison and Limit Comparison Tests — Question 6

*Question 6 Consider $\displaystyle\sum_{n=1}^{\infty}\frac1{n^2-n+1}$.

1. Prove useful upper and lower quadratic bounds for the denominator.

2. Convert them into bounds for the summand, noting reciprocal inequality directions.

3. Classify the series and confirm by limit comparison.

*Question 6 – Solution

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 $n^2/2\le n^2-n+1\le n^2$.

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.

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