Does $\sum_{n=1}^{\infty}\frac{n-1}{n^2}$ converge or diverge?

Question

Is my logic OK?

$a_{n}=\frac{n-1}{n^2}$

$\frac{1}{n} \leq b_{n}=\frac{n-\frac{n}{2}}{n^2}=\frac{n}{2n^2}=\frac{1}{2n} \leq a_{n}=\frac{n-1}{n^2}$

and there for the initial series diverges.

Answer 1

You have the right idea, but the wrong execution. Your chain of inequalities is wrong for every $n$, sadly - the error is at the beginning, involving $b_n$.

To fix it, you could note that

$$\frac{1}{n^2} \le \frac 1 {2n}$$

for all $n \ge 2$, so that

$$\frac{n - 1}{n^2} = \frac1 n - \frac{1}{n^2} \ge \frac 1 n - \frac 1 {2n} = \frac 1 2 \cdot \frac 1 n$$

Answer 2

$$\sum_{n=1}^\infty\frac{n-1}{n^2}=\sum_{n=1}^\infty\frac{1}{n}-\sum_{n=1}^\infty\frac{1}{n^2}$$ is the sum of a divergent series and a convergent series so it's a divergent series.

Answer 3

You're attempting to use the comparison test, but the inequalities in the comparison test can sometimes be troublesome. I prefer the limit comparison test.

Letting $a_n=1/n$ and $b_n=(n-1)/n^2$ gives $$ \lim_{n\to\infty}\frac{a_n}{b_n}=\lim_{n\to\infty}\frac{n^2}{n(n-1)}=\lim_{n\to\infty}\frac{n}{n-1}=1 $$ Hence the limit comparison test implies either both $\sum a_n$ and $\sum b_n$ converge or both diverge. Of course, $\sum a_n$ is the harmonic series so both series diverge.