- #1
3trQN
- 337
- 1
I Have a small problem.
I know the following:
The series:
[tex]
S_{n} = 1 + 2 + 3 + 4 + 5 = \sum_{i=1}^{n=5}i
[/tex]
[tex]
2S_{n} = 1 + 2 + 3 + 4 + 5 + 1 + 2 + 3 + 4 + 5 = n(n+1) = n^{2} + n
[/tex]
Therefor:
[tex]
S_{n} = \frac{2S_{n}}{2} = \frac{n^{2} + n}{2}
[/tex]
[tex]
\sum_{i=1}^{n=5}i = \frac{n^{2} + n}{2} = \frac{25 + 5}{2} = 15
[/tex]
Ok, that's simple.
But i was attempting to do the same with the factorials, where:[tex]
ln(n!) = ln(5!) = ln(1) + ln(2) + ... + ln(5) = \sum_{i=1}^{n=5}log(i)
[/tex]
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.
I know the following:
The series:
[tex]
S_{n} = 1 + 2 + 3 + 4 + 5 = \sum_{i=1}^{n=5}i
[/tex]
[tex]
2S_{n} = 1 + 2 + 3 + 4 + 5 + 1 + 2 + 3 + 4 + 5 = n(n+1) = n^{2} + n
[/tex]
Therefor:
[tex]
S_{n} = \frac{2S_{n}}{2} = \frac{n^{2} + n}{2}
[/tex]
[tex]
\sum_{i=1}^{n=5}i = \frac{n^{2} + n}{2} = \frac{25 + 5}{2} = 15
[/tex]
Ok, that's simple.
But i was attempting to do the same with the factorials, where:[tex]
ln(n!) = ln(5!) = ln(1) + ln(2) + ... + ln(5) = \sum_{i=1}^{n=5}log(i)
[/tex]
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.
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