# Series and sequences

• Nov 28th 2008, 03:25 AM
tsal15
Series and sequences
I'm having problem with this question.

Do I use the ratio test for this infinite series? if so, can you show a few steps to get me going, or if not, what should I use?

$\displaystyle \sum\limits_{n=1}^{\infty} \frac{2^n - n^5}{4^n+2n^2+3n}$

• Nov 28th 2008, 03:53 AM
mr fantastic
Quote:

Originally Posted by tsal15
I'm having problem with this question.

Do I use the ratio test for this infinite series? if so, can you show a few steps to get me going, or if not, what should I use?

$\displaystyle \sum\limits_{n=1}^{\infty} \frac{2^n - n^5}{4^n+2n^2+3n}$

You have $\displaystyle \sum\limits_{n=1}^{\infty} \frac{2^n}{4^n+2n^2+3n} - \sum\limits_{n=1}^{\infty} \frac{n^5}{4^n+2n^2+3n}$.

Now note:

1. $\displaystyle \frac{2^n}{4^n+2n^2+3n} < \frac{2^n}{4^n} = \left(\frac{1}{2}\right)^n$.

So the first sum converges by the comparison test.

2. $\displaystyle \frac{n^5}{4^n+2n^2+3n} < \frac{n^5}{4^n}$ and $\displaystyle \sum\limits_{n=1}^{\infty} \frac{n^5}{4^n}$ converges by the ratio test.

So the second sum converges.
• Nov 28th 2008, 03:58 AM
tsal15
Quote:

Originally Posted by mr fantastic
You have $\displaystyle \sum\limits_{n=1}^{\infty} \frac{2^n}{4^n+2n^2+3n} - \sum\limits_{n=1}^{\infty} \frac{n^5}{4^n+2n^2+3n}$.

Now note:

1. $\displaystyle \frac{2^n}{4^n+2n^2+3n} < \frac{2^n}{4^n} = \left(\frac{1}{2}\right)^n$.

So the first sum converges by the comparison test.

2. $\displaystyle \frac{n^5}{4^n+2n^2+3n} < \frac{n^5}{4^n}$ and $\displaystyle \sum\limits_{n=1}^{\infty} \frac{n^5}{4^n}$ converges by the ratio test.

So the second sum converges.

Thank you for the quick reply Mr. Fantastic. You truly are fantastic (Clapping).
• Nov 28th 2008, 07:49 AM
tsal15
Quote:

Now note:

1. $\displaystyle \frac{2^n}{4^n+2n^2+3n} < \frac{2^n}{4^n} = \left(\frac{1}{2}\right)^n$.

So the first sum converges by the comparison test.
Oh Mr. Fantastic I should've asked you this earlier...I was just soo excited someone actually replied to my post... you see i haven't had much help for a while on mathhelp...but any who. how did you know to compare that sum with $\displaystyle \frac{2^n}{4^n}$?

Quote:

2. $\displaystyle \frac{n^5}{4^n+2n^2+3n} < \frac{n^5}{4^n}$ and $\displaystyle \sum\limits_{n=1}^{\infty} \frac{n^5}{4^n}$ converges by the ratio test.

So the second sum converges.
I believe you've used the comparison test, again, after finding out $\displaystyle \frac{n^5}{4^n}$ converges by the ratio test? Similarly, how did you know to use $\displaystyle \frac{n^5}{4^n}$

Thank you Mr. Fantastic :)
• Nov 28th 2008, 01:23 PM
mr fantastic
Quote:

Originally Posted by tsal15
Oh Mr. Fantastic I should've asked you this earlier...I was just soo excited someone actually replied to my post... you see i haven't had much help for a while on mathhelp...but any who. how did you know to compare that sum with $\displaystyle \frac{2^n}{4^n}$?

I believe you've used the comparison test, again, after finding out $\displaystyle \frac{n^5}{4^n}$ converges by the ratio test? Similarly, how did you know to use $\displaystyle \frac{n^5}{4^n}$ Mr F says: Yes, I did.

Thank you Mr. Fantastic :)

In both cases it's a matter of guessing whether the series is convergent or divergent and then trying to construct an appropriate series to use in the comparison. The key is a lot of practice to develop experience - after a while you get to 'see' (sometimes) what the series will be that you need to use.
• Nov 28th 2008, 01:30 PM
tsal15
Quote:

Originally Posted by mr fantastic
In both cases it's a matter of guessing whether the series is convergent or divergent and then trying to construct an appropriate series to use in the comparison. The key is a lot of practice to develop experience - after a while you get to 'see' (sometimes) what the series will be that you need to use.

hmmm, guessing i can get used to that...hehehe. but yes i do agree with, practice will help.

thanks Mr. Fantastic :)