# Help With Series

• Oct 24th 2012, 01:45 PM
Preston019
Help With Series
Determine whether the series ∑_(n=1)^(infinity) ((squareroot(n)+ln(n))/((n^2)+1)) converges or diverges. clearly identigy any test(s) you are using and label all relevant data and/or properties.

I have no idead where to start or what tests to use.
• Oct 24th 2012, 02:18 PM
benb89
Re: Help With Series
It can be shown that $\ln(n) \leq \sqrt{n}$

Can you then find a bound on each term of your series?

Hint - think comparison test and p-test to arrive at your final answer!
• Oct 24th 2012, 02:46 PM
Preston019
Re: Help With Series
Quote:

Originally Posted by benb89
It can be shown that $\ln(n) \leq \sqrt{n}$

Can you then find a bound on each term of your series?

Hint - think comparison test and p-test to arrive at your final answer!

Are you saying compare it to squareroot(n)/n?
• Oct 24th 2012, 03:01 PM
benb89
Re: Help With Series
So,

$\frac{\ln n + \sqrt{n}}{n^2 + 1}\leq \frac{\sqrt{n} + \sqrt{n}}{n^2 + 1}$

Can you go forward from here?
• Oct 24th 2012, 03:04 PM
Plato
Re: Help With Series
NO!
$\frac{\sqrt{n}+\ln(n)}{n^2+1}\le\frac{2\sqrt{n}}{n ^2+1}<\frac{2}{n^{3/2}}$.
The last is a p-series that converges.
• Oct 24th 2012, 03:05 PM
Preston019
Re: Help With Series
Quote:

Originally Posted by benb89
So,

$\frac{\ln n + \sqrt{n}}{n^2 + 1}\leq \frac{\sqrt{n} + \sqrt{n}}{n^2 + 1}$

Can you go forward from here?

I'm not sure. I'm using the comparison test right? Or could I use the limit comparison test?
• Oct 24th 2012, 03:08 PM
Preston019
Re: Help With Series
Quote:

Originally Posted by Plato
NO!
$\frac{\sqrt{n}+\ln(n)}{n^2+1}\le\frac{2\sqrt{n}}{n ^2+1}<\frac{2}{n^{3/2}}$.
The last is a p-series that converges.

Oh! Okay. So am I going to have to do multiple comparison tests?
• Oct 24th 2012, 03:21 PM
Plato
Re: Help With Series
Quote:

Originally Posted by Preston019
Oh! Okay. So am I going to have to do multiple comparison tests?

No you don't.
Just say $\sum\limits_{k = 1}^\infty {\frac{2}{{\sqrt[2]{{n^3 }}}}}$ converges.
• Oct 24th 2012, 03:33 PM
Preston019
Re: Help With Series
Quote:

Originally Posted by Plato
No you don't.
Just say $\sum\limits_{k = 1}^\infty {\frac{2}{{\sqrt[2]{{n^3 }}}}}$ converges.

Okay. And where did you get ln(n) is less than or equal to the square root of n?