# A textbook question

1. Nov 30, 2005

### Swatch

In my textbook there is an example that shows determination whether a series converges or diverges using the limit comparison test.
"
The series is (1+n *ln(n)) / (n^2 + 5)
For n large, we expect an to behave like (n*ln(n))/n^2 = (ln(n))/n, which is greater than 1/n for n>= 3, so we take bn = 1/n.
"

My question is why can I use 1/n as an comparison series, I guess it's because its smaller than (ln(n))/n for large n but I'm not sure how that should validate the choice.

Could someone please explain to me?

2. Nov 30, 2005

### HallsofIvy

Staff Emeritus
"I guess it's because its smaller than (ln(n))/n for large n "

Yes, that's the whole point- together with the fact that $\Sigma \frac{1}{n}$ does not converge. The "Limit Comparison Theorem" says that if $a_n\leq b_n$ for large enough n, and $\Sigma a_n$ does not converge, then $\Sigma b_n$ does not converge.

3. Nov 30, 2005

### math-chick_41

they used 1/n because they had a hunch that the series diverges. so when you do the limit comparison test you calculate lim an/bn and if the answer is a finite number greater than zero then either both of your series converge or both diverge. since 1/n is a known divergent series the series you compared it to must also diverge.
it is good test to know for an exam because it is so easy and doesnt require a bunch of inequalities and lots of thinking. if you suspesct your series an to converge then just pick a series bn that converges and calculate lim an/bn if its finite and bigger than zero then you are done.

4. Nov 30, 2005

Thanks.