# How to chose the resembling series in the comparison test

I am having some difficulties choosing the resembling series for the comparison test.

This problem is an example:

$$a_n=\sum_{n=1}^\infty \left(\frac{5}{n}-\frac{5}{n^2}\right)$$

The way I think about it is: The expression can be manipulated to $\left(\dfrac{5n-5}{n^2}\right)$, which is similar to the harmonic series

$$\sum_{n=1}^\infty \frac{1}{n},$$

but the solutions manual suggests using

$b_n=\sum_{n=1}^\infty\left(\dfrac{1}{2n}\right)$, and I am wondering if someone can explain me the logic behind their choice?

Also, as far as I can see, the terms $a_n > b_n$, which I thought meant that the test was inconclusive...

If you can show $a_n > b_n$, and $\sum b_n$ diverges, then by comparison test, your series diverges. $\sum \frac{1}{2n}$ or $\sum \frac{1}{n}$ doesn't make much difference to the convergence result.
Another way to show the divergence is that if $\sum(\frac{5}{n}-\frac{5}{n^2})$ converges and since $\sum \frac{5}{n^2}$ converges, you will get $\sum \frac{5}{n}=\sum(\frac{5}{n}-\frac{5}{n^2})+ \sum \frac{5}{n^2}$ converges, which is impossible. So the series must diverge.