1. ## Sequence Help

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

2. Try [shift]+1

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

3. 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?

4. Expand the factorials. Only two terms differ.

5. Originally Posted by TKHunny
Expand the factorials. Only two terms differ.
So when it is fully evaluated it would be 1 / (4n^2 + 2n)

6. "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.)

7. 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?

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

9. 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 $\frac{1}{2(1)} - \frac{1}{2(1)+1}\;=\;\frac{1}{2}-\frac{1}{3}$

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

10. Originally Posted by TacticalPro
Never noticed that, how did you get it? and how does that lead to the sum?
$\frac{1}{2n(2n+1)}=\frac{1}{2n}- \frac{1}{2n+1}$

11. 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

12. This serie is not telescopic

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

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

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