Sequence Help

**TacticalPro** · June 6th 2011, 06:21 PM

I'm studying for my final and my teacher gave me a worksheet. I have two sequence problems I do not understand at all. If anyone could please help I'd appreciate it!

1. Find the sum of the sequence, note i=symbol for factorial, I couldn't find a factorial symbol

:

((2n-1)i) / ((2n+1)i)

2. find sum of sequence (That's exactly how written, I have no idea...) :

2, -1/2 + 1/8 - ... 1/2048

**TKHunny** · June 6th 2011, 06:36 PM

Try [shift]+1

Have you noticed that $\frac{(2n-1)!}{(2n+1)!} = \frac{1}{2n(2n+1)}$ ?

**TacticalPro** · June 6th 2011, 06:47 PM

Originally Posted by TKHunny

Try [shift]+1

Have you noticed that $\frac{(2n-1)!}{(2n+1)!} = \frac{1}{2n(2n+1)}$ ?

Never noticed that, how did you get it? and how does that lead to the sum?

**TKHunny** · June 6th 2011, 06:50 PM

Expand the factorials. Only two terms differ.

**TacticalPro** · June 6th 2011, 06:53 PM

Originally Posted by TKHunny

Expand the factorials. Only two terms differ.

So when it is fully evaluated it would be 1 / (4n^2 + 2n)

**TKHunny** · June 6th 2011, 07:03 PM

"fully evaluated" means nothing.

Finding a parsimonious expression for a sum can be an art. Wonderful accolades have been awarded because of such discoveries. Other than self-awesomeness, though, you need a bag of references. In this case, notice further that:

$\frac{1}{2n(2n+1)}\;=\;\frac{1}{2n} - \frac{1}{2n+1}$

Write out the expression for n = 1, 2, and 3 and see if you notice anything. (Resist the temptation to add anything.)

**TacticalPro** · June 6th 2011, 07:09 PM

Originally Posted by TKHunny

"fully evaluated" means nothing.

Finding a parsimonious expression for a sum can be an art. Wonderful accolades have been awarded because of such discoveries. Other than self-awesomeness, though, you need a bag of references. In this case, notice further that:

$\frac{1}{2n(2n+1)}\;=\;\frac{1}{2n} - \frac{1}{2n+1}$

Write out the expression for n = 1, 2, and 3 and see if you notice anything. (Resist the temptation to add anything.)

I don't quite understand what you're getting at. I made a mistake in the original post, it doesn't want the sum of the expression, but rather it says to evaluate it.

also, anything on the second question?

**TheChaz** · June 6th 2011, 07:23 PM

2, -1/2 + 1/8 - ... +1/2048
This looks like a geometric sequence with first term "2" and ratio "-1/4"...

**TKHunny** · June 6th 2011, 07:35 PM

You probably should differentiate between a sequence (a list) and a series (add things up).

Did you list the terms associated with n = 1, 2, and 3? Really, it will be quite enlightening. You will fail this course if you can't list three terms of a series.

n = 1 leads to $\frac{1}{2(1)} - \frac{1}{2(1)+1}\;=\;\frac{1}{2}-\frac{1}{3}$

How about n = 2?

Note: That ratio on #2 looks like "-1/4" and it appears to be finite.

**abhishekkgp** · June 6th 2011, 08:05 PM

Originally Posted by TacticalPro

Never noticed that, how did you get it? and how does that lead to the sum?

$\frac{1}{2n(2n+1)}=\frac{1}{2n}- \frac{1}{2n+1}$

**skeeter** · June 7th 2011, 04:32 AM

Originally Posted by TKHunny

Have you noticed that $\frac{(2n-1)!}{(2n+1)!} = \frac{1}{2n(2n+1)}$ ?

Never noticed that, how did you get it? and how does that lead to the sum?

Telescoping series - Wikipedia, the free encyclopedia

**Renji Rodrigo** · June 7th 2011, 05:43 AM

This serie is not telescopic

$\sum^{\infty}_{k=1}\frac{1}{2k} -\frac{1}{2k+1}=\sum^{\infty}_{k=0}\frac{1}{2k+2} -\frac{1}{2k+3}=\sum^{\infty}_{k=0} \int^{1}_{0} x^{2k+1}(1-x)dx =$

$=\int^{1}_{0} (1-x)x\sum^{\infty}_{k=0} x^{2k}dx=\int^{1}_{0}\frac{x(1-x)}{1-x^2}dx = 1 - \ln (2)$

$=\sum^{\infty}_{k=1}\frac{1}{(2k)(2k+1)} = 1 - \ln (2) .$

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