I'm studying summation. Everything I know so far is that:
$\sum_{i=1}^n\ k = \frac{n(n+1)}{2}\ $
$\sum_{i=1}^{n}\ k^2 = \frac{n(n+1)(2n+1)}{6}\ $
$\sum_{i=1}^{n}\ k^3 = \frac{n^2(n+1)^2}{4}\ $
Unfortunately, I can't find neither on my book nor on the internet what the result of:
$\sum_{i=1}^n\log i$.
$\sum_{i=1}^n\ln i$.
is.
Can you help me out?
$\endgroup$ 22 Answers
$\begingroup$By using the fact that $$\log a + \log b = \log ab $$ then
$$ \sum^n \log i = \log (n!) $$
$$ \sum^n \ln i = \ln (n!) $$
$\endgroup$ 1 $\begingroup$Since $\log(A)+\log(B)=\log(AB)$, then $\sum_{i=1}^n\log(i)=\log(n!)$. I'm not sure if this helps a lot since you have changed a summation of $n$ terms into a product of $n$ factors, but it's something.
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