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Misael Matthews

Misael Matthews

Answered question

2022-06-14

Best approximation for k = 2 n ln ln k ?
I need the best approximation for k = 2 n ln ( ln ( k ) ) . Any suggestion or hint is welcomed.
I derived n ln ( ln ( n ! ) n ) so is there any better one ?

Answer & Explanation

Josie Stephenson

Josie Stephenson

Beginner2022-06-15Added 20 answers

Euler-Maclaurin series:
k = 2 n ln ( ln ( k ) ) = C + 2 n ln ( ln ( t ) )   d t + 1 2 ln ( ln ( n ) ) + 1 12 n ln ( n ) 1 360 n 3 ln ( n ) 1 240 n 3 ln ( n ) 2 1 360 n 3 ln ( n ) 3 + O ( 1 / n 5 )
for some constant C, where
2 n ln ( ln ( t ) )   d t = n ln ( ln ( n ) ) 2 ln ( ln ( 2 ) ) 2 n d t ln ( t ) = n ln ( ln ( n ) ) L i ( n ) 2 ln ( ln ( 2 ) ) + L i ( 2 )
Numerically it appears C .2412388
Roland Manning

Roland Manning

Beginner2022-06-16Added 5 answers

You can use the Euler-Maclaurin Sum Series. The first few terms are
2 n log ( log ( x ) ) d x + 1 2 log ( log ( n ) ) + C + 1 12 n log ( n ) +
We won't worry about any terms past the first since we need to approximate the integral asymptotically, and the terms in that expansion are bigger than log ( log ( n ) ).
log ( log ( x ) ) d x = log ( u ) d e u = e u log ( u ) e u u d u = e u log ( u ) e u u e u u 2 d u = e u log ( u ) e u u e u u 2 2 e u u 3 d u = e u log ( u ) e u u e u u 2 2 e u u 3 6 e u u 4 d u = e u log ( u ) e u u e u u 2 2 e u u 3 6 e u u 4 24 e u u 5 d u
From this, we get the asymptotic expansion
k = 2 n log ( log ( k ) ) n log ( log ( n ) ) n log ( n ) ( 1 + 1 log ( n ) + 2 log ( n ) 2 + 6 log ( n ) 3 + )

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