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2007-07-21 11:40:47 · 3 answers · asked by Amir 1 in Science & Mathematics Mathematics

3 answers

Find the sum for the series.

I assume we are working strictly with integers and factorials. If so, the series is not infinite.

For odd integers n, we have

[nC0] - [(n-1)C1] + [(n-2)C2] ... [(n+1)/2)C(n-1)/2]

It stops at the last term shown because you can't choos more than you have to choose from.

For even integers n, we have

[nC0] - [(n-1)C1] + [(n-2)C2] ... [(n/2)C(n/2)]

It stops at the last term shown because you can't choos more than you have to choose from.
_____________

The sums are suprising. For starting n we have the following sums.

n.. .. sum
01 ... 1
02 ... 0
03 ... -1
04 ... -1
05 ... 0
06 ... 1
07 ... 1
08 ... 0
09 ... -1
10 ... -1
11 ... 0
12 ... 1
etc.
________

2007-07-21 12:38:55 · answer #1 · answered by Northstar 7 · 0 0

Well, let's examine what happens when we take the i-th term in the series, then make i very large:

The sequence looks like this (in i):
(-1)^i * Com[n-i, i]
where Com[x, y] = x! / (y! * (x-y)!)

So, the i-th term is
(-1)^i * (n-i)! / (i!*(n - 2*i)!)

When n is any integer, the series is undefined, because you'll have a negative integer factorial in either the numerator or denominator. The Euler gamma function defines the factorial for values that aren't positive integers, but it blows up at all negative integers.

When n is a non-integer, the series grows without bound. So, the sum of the entire series is infinity for any non-integer value.

Here are the first few terms with n = 1.5, starting with i = 0, and going to i = 10:

1, -0.5, 0.375, 2.1875, 9.02344, 35.1914, 135.319, 518.724, 1989.42, 7643.13, 29426.1

As you can, see the terms grow larger and larger, so the series blows up to infinity as i goes to infinity.

You can read up on the Euler gamma function here:
http://mathworld.wolfram.com/GammaFunction.html

2007-07-21 12:00:00 · answer #2 · answered by lithiumdeuteride 7 · 0 0

Just by observation, the series follows the pattern
1,1,0,-1,-1,0,1,1,0 etc
for n = 0,1,2,...

f(n+6) = f(n)
f(0) = f(1) = 1
f(2) = f(5) = 0
f(3) = f(4) = -1

2007-07-21 17:11:59 · answer #3 · answered by Dr D 7 · 0 0

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