Factorials & Series - im a little lost here

[3trQN]

I Have a small problem.

I know the following:

The series:
$$S_{n} = 1 + 2 + 3 + 4 + 5 = \sum_{i=1}^{n=5}i$$

$$2S_{n} = 1 + 2 + 3 + 4 + 5 + 1 + 2 + 3 + 4 + 5 = n(n+1) = n^{2} + n$$

Therefor:
$$S_{n} = \frac{2S_{n}}{2} = \frac{n^{2} + n}{2}$$

$$\sum_{i=1}^{n=5}i = \frac{n^{2} + n}{2} = \frac{25 + 5}{2} = 15$$

Ok, that's simple.
But i was attempting to do the same with the factorials, where:$$ln(n!) = ln(5!) = ln(1) + ln(2) + ... + ln(5) = \sum_{i=1}^{n=5}log(i)$$

Is it possible to apply the same method, or something similar here? I did try a few but got a bit cheesed off.

In the first case, you just duplicate and reverse the sequence and add to get a constant sequence.

 

[Robokapp]

$$ln(n!) = ln(5!) = ln(1) + ln(2) + ... + ln(5) = \sum_{i=1}^{n=5}ln(i)$$

Don't you mean that? Sum of Ln(i) ?

$$2\sum_{i=1}^{n=5}i$$ can also be written as $$\sum_{i=1}^{n=5}2i$$ if I'm not mistaking. Your algebration wrote it as $$\sum_{i=1}^{n=5}i + \sum_{i=1}^{n=5}i$$ which is correct but not necessarily useful.

Now...If I recall the Logaritmic properties, Log(a)+Log(b)=Log(ab)

So $$ln(n!)= Ln(1*2*3*4*5...n) = Ln(1)+Ln(2)+Ln(3)...+Ln(n)=\sum_{i=1}^{n=oo}ln(i)$$

From what I can tell you are correct. Well except n=5 part. If you have
$$ln(n!)$$ without any other condition it will equal $$\sum_{i=1}^{n=oo}ln(i)$$

 

[3trQN]

Yes, thanks.

Now...If I recall the Logaritmic properties, Log(a)+Log(b)=Log(ab)

Erm, yeah, working the other way though :P

The idea is, one fast method for calculating the sum of the positive integers to a given value, say 10, is to take 0.5(10^2 + 10).

Well i was wondering if there is a shortcut for factorials in a similar manner.

Its late, and the coffee has stopped working...

 

[Robokapp]

The idea is, one fast method for calculating the sum of the positive integers to a given value, say 10, is to take 0.5(10^2 + 10).

That is true...but all I got in my Pre-Calculus about sums is some basic ways to do them, no explanations.

(n^2+n)/2 if you have a sum of an unknown with power of 1 starting at 1, (n^2-n)/2 if it starts at zero I think...and some formulas for situations when you make a Sum of...n^2 and n^3 etc...But never why.

They showed some substraction between terms, double substraction if its quadratic until the difference is the same...something about a 3x4 matrix for quadratics...If I stretch my brain i may be able to reproduce it but it didn't explain most of anything. That is one area of Mathematics that I'd like to revisit in the future.

 

[rbj]

there is no exact formula, but there is a very good approximation for $\log(n!)$ that gets better as $n$ gets large. it's called Stirling's approximation and is:

$$\frac{1}{12 n + 1} < \log \left( \frac{n!}{\sqrt{2 \pi n}} \right) - n \left( \log(n) - 1 \right)< \frac{1}{12 n}$$

or

$$n \left( \log(n) - 1 \right) + \frac{\log(2 \pi n)}{2} + \frac{1}{12 n + 1} < \log(n!) < n \left( \log(n) - 1 \right) + \frac{\log(2 \pi n)}{2} + \frac{1}{12 n}$$

i would suggest to Google or Wikipedia it to find out why this is true.

 

[shmoe]

You might want to look up Euler-Maclaurin summation.

 

[3trQN]

Thanks, got a fresh day and a new cup of coffee. Ill read up on those two topics.

 

[Ulagatin]

You can get back to the nth factorial with this representation:

$$e^{\sum_{k=1}^n ln(k)}$$

Not sure if this is at all helpful to you, but there's a statement that returns you to n!

Try it out, I assure you it works, except if you let k= 0 as ln(0) is undefined.