# Comparison & Limit Comparison test for series

• Mar 25th 2009, 10:30 AM
mollymcf2009
Comparison & Limit Comparison test for series
Can someone see if I am doing this the right way? Or if there is an EASIER way to do this?

Determine whether the series converges or diverges.

$\sum^{\infty}_{n=1} \frac{n^2 + 1}{n^2 \sqrt{n}}$

*Pulled out my variables:
$\sum \frac{n^{\frac{4}{2}}}{n^{\frac{5}{2}}}$

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

So,

$\lim_{n \rightarrow \infty} \frac{n^{\frac{4}{2}}}{n^{\frac{5}{2}}} \cdot \frac{1}{n^{\frac{1}{2}}}$

$\rightarrow \lim_{n\rightarrow \infty} \frac{n^2}{n^3}$

$\rightarrow \lim_{n\rightarrow \infty} \frac{1}{n} = 0$

BUT, because $n^{\frac{1}{2}}$ is DIVERGENT then this series is also divergent. Right?

This stuff is SO abstract! Any help is GREATLY appreciated! Thanks! Molly
• Mar 25th 2009, 11:42 AM
matheagle
Quote:

Originally Posted by mollymcf2009
Can someone see if I am doing this the right way? Or if there is an EASIER way to do this?

Determine whether the series converges or diverges.

$\sum^{\infty}_{n=1} \frac{n^2 + 1}{n^2 \sqrt{n}}$

*Pulled out my variables:
$\sum \frac{n^{\frac{4}{2}}}{n^{\frac{5}{2}}}$

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

So,

$\lim_{n \rightarrow \infty} \frac{n^{\frac{4}{2}}}{n^{\frac{5}{2}}} \cdot \frac{1}{n^{\frac{1}{2}}}$

$\rightarrow \lim_{n\rightarrow \infty} \frac{n^2}{n^3}$

$\rightarrow \lim_{n\rightarrow \infty} \frac{1}{n} = 0$

BUT, because $n^{\frac{1}{2}}$ is DIVERGENT then this series is also divergent. Right?

This stuff is SO abstract! Any help is GREATLY appreciated! Thanks! Molly

you should DIVIDE the terms not multiply
It's correct to comare your orginal series to $\sum^{\infty}_{n=1} \frac{1} {\sqrt{n}}=\infty$ since p=1/2.

Then you take the RATIO of the two terms.

$\lim_{n\to\infty}{ {n^2 + 1\over n^2 \sqrt{n}}\over {1\over \sqrt{n}}}=\lim_{n\to\infty} {n^2 + 1\over n^2 }=1$

ANOTHER point, the sequence $\frac{1} {\sqrt{n}}$ converges to zero

The series $\sum^{\infty}_{n=1} \frac{1} {\sqrt{n}}$ diverges (goes) to infinity.

Do not confuse a sequence with a sum.
• Mar 25th 2009, 12:28 PM
mollymcf2009
Quote:

Originally Posted by matheagle
you should DIVIDE the terms not multiply
It's correct to comare your orginal series to $\sum^{\infty}_{n=1} \frac{1} {\sqrt{n}}=\infty$ since p=1/2.

Then you take the RATIO of the two terms.

$\lim_{n\to\infty}{ {n^2 + 1\over n^2 \sqrt{n}}\over {1\over \sqrt{n}}}=\lim_{n\to\infty} {n^2 + 1\over n^2 }=1$

ANOTHER point, the sequence $\frac{1} {\sqrt{n}}$ converges to zero

The series $\sum^{\infty}_{n=1} \frac{1} {\sqrt{n}}$ diverges (goes) to infinity.

Do not confuse a sequence with a sum.

I did divide, just multiplied by reciprocal. But, I didn't get $\frac{1}{n^{1/2}}$ I got just $n^{1/2}$ and $n^{1/2}$ goes to infinity as n goes to infinity right? Hence diverges, so because it diverges so does the original series?
• Mar 25th 2009, 01:06 PM
o_O
$\frac{n^2 + 1}{n^2 \sqrt{n}} = \frac{n^2 + 1}{n^{\frac{5}{2}} } {\color{red} \ > \ } \frac{n^2}{n^{\frac{5}{2}}} = \frac{1}{n^{\frac{1}{2}}}$

So by the comparison test ...
• Mar 25th 2009, 05:00 PM
matheagle
Nope, you have a couple of error.

You set a sum equal to a limit, and the alegbra is wrong too.

You should have ${n^{4/2}\over n^{5/2}} ={1\over n^{1/2}}$.