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)!))

Jacoby Erickson 2022-10-14 Answered
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 ) ! )
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Answers (1)

faux0101d
Answered 2022-10-15 Author has 21 answers
Note that
log ( 1 ) + log ( 2 ) + log ( 3 ) + . . . . log ( n 1 ) + log ( n ) = log ( n ! )
and
log ( a ) + log ( a + 1 ) + log ( a + 2 ) + . . . . log ( n 1 ) + log ( n ) = log i = a n i
Note: You need the property
log ( a b ) = log ( a ) + log ( b ) .
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