Is this summation solvable? S_n=sum_(i=2)^n log_i(n) Should I use the derivative of log_i(n)?

Harper George 2022-09-30 Answered
Is this summation solvable? S n = i = 2 n log i ( n )
Is it possible to solve a summation with a variable base of log?
S n = i = 2 n log i ( n )
Should I use the derivative of log i ( n )
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Answers (2)

Collin Gilbert
Answered 2022-10-01 Author has 11 answers
While I don't believe there is a nice closed form for S n , you can write the sum in terms of known functions and constants up to a very small error. Specifically,
i = 2 N log i ( N ) = li(N) log N + C log N + O ( 1 ) ,
where li ( N ) is the logarithmic integral and C is a constant equal to
C = 1 log 2 + 2 { x } x log 2 x d x .
Proof: Writing
i = 2 N log i ( N ) = log N i = 2 N 1 log i ,
our goal is then to find an asymptotic for the sum of 1 / log i. Writing this as a Riemann Stieltjies integral we have
i = 2 N 1 log i = 2 N + 1 log x d [ x ] = 2 N 1 log x d x 2 N + 1 log x d { x } .
By integration by parts,
2 N + 1 log x d { x } = { x } log x | x = 2 x = N + + 2 N { x } x log 2 x d x
= 1 log 2 + 2 { x } x log 2 x d x N { x } x log 2 x d x ,
and since
N { x } x log 2 x d x = O ( 1 log N ) ,
we have that
i = 2 N 1 log i = li ( N ) + C + O ( 1 log N )
where li ( N ) is the logarithmic integral and
C = 1 log 2 + 2 { x } x log 2 x d x .
Thus it follows that
i = 2 N log i ( N ) = li(N) log N + C log N + O ( 1 ) .
(Note that the asymptotic is then i = 2 N log i ( N ) N .)
Remark: In fact, we could apply integration by parts again to work out the O ( 1 ) term exactly and evaluate the sum up to an error of O ( 1 N ) . This general process of writing the sum k N f ( k ) as a series whose main term is 1 N f ( x ) d x is known as Euler-Maclaurin summation.
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deksteenk9
Answered 2022-10-02 Author has 1 answers
Since log i n = log n log i , we have
S n = i = 2 n log n log i = log n i = 2 n 1 log i
and we have by Euler-McLaurin summation formula
i = 2 n 1 log i = 2 n d x log x + log 2 n + O ( 1 log n ) ,
which leads us to
S n = Li ( n ) log n + 1 2 log 2 n + log 2 log n + O ( 1 ) .
Hope this helps.
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