# What's an intuitive way to compute summation of this series? What's an intuitive way to compute log(1)+log(2)+log(3)+⋯+log(n−1)+log(n) or for n>a log(a)+log(a+1)+log(a+2)+⋯+log(n−1)+log(n) I know the answer for general case is log ((n!)/((a-1)!))

What's an intuitive way to compute summation of this series?
What's an intuitive way to compute
$\mathrm{log}\left(1\right)+\mathrm{log}\left(2\right)+\mathrm{log}\left(3\right)+\cdots +\mathrm{log}\left(n-1\right)+\mathrm{log}\left(n\right)$
or
for $n>a$
$\mathrm{log}\left(a\right)+\mathrm{log}\left(a+1\right)+\mathrm{log}\left(a+2\right)+\cdots +\mathrm{log}\left(n-1\right)+\mathrm{log}\left(n\right)$
I know the answer for general case is
$\mathrm{log}\left(\frac{n!}{\left(a-1\right)!}\right)$
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

faux0101d
Note that
$\mathrm{log}\left(1\right)+\mathrm{log}\left(2\right)+\mathrm{log}\left(3\right)+....\mathrm{log}\left(n-1\right)+\mathrm{log}\left(n\right)=\mathrm{log}\left(n!\right)$
and
$\mathrm{log}\left(a\right)+\mathrm{log}\left(a+1\right)+\mathrm{log}\left(a+2\right)+....\mathrm{log}\left(n-1\right)+\mathrm{log}\left(n\right)=\mathrm{log}\prod _{i=a}^{n}i$
Note: You need the property
$\mathrm{log}\left(ab\right)=\mathrm{log}\left(a\right)+\mathrm{log}\left(b\right).$