Series and sequences

**tsal15** · November 28th 2008, 03:25 AM

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?

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

Thanks in advance

**mr fantastic** · November 28th 2008, 03:53 AM

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?

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

Thanks in advance

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

Now note:

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

So the first sum converges by the comparison test.

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

So the second sum converges.

**tsal15** · November 28th 2008, 03:58 AM

Originally Posted by mr fantastic

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

Now note:

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

So the first sum converges by the comparison test.

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

So the second sum converges.

Thank you for the quick reply Mr. Fantastic. You truly are fantastic

.

**tsal15** · November 28th 2008, 07:49 AM

Now note:

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

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 $\frac{2^n}{4^n}$ ?

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

So the second sum converges.

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

Thank you Mr. Fantastic

**mr fantastic** · November 28th 2008, 01:23 PM

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 $\frac{2^n}{4^n}$ ?

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

**tsal15** · November 28th 2008, 01:30 PM

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

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