Convergence of series minus logarithm im trying to solve this problem since two, three days..

Makenna Booker

Makenna Booker

Answered question

2022-07-15

Convergence of series minus logarithm
im trying to solve this problem since two, three days.. Is there someone who can help me to solve this problem step by step. I really want to understand & solve this!
S h o w     β [ 0 , 1 ] , s o   t h a t . . lim n ( k = 2 n 1 k log ( k ) log ( log ( n ) ) ) = β  
The sum looks like the harmonic series.
My thoughts were to compare this sum with an integral, the lower and the upper riemann-sum, to get an inequation.

Answer & Explanation

iljovskint

iljovskint

Beginner2022-07-16Added 18 answers

Comparing sum and integral, we get
1 n log ( n ) + 2 n d x x log ( x ) k = 2 n 1 k log ( k ) (1) 1 2 log ( 2 ) + 1 3 log ( 3 ) + 3 n d x x log ( x )
Thus,
1 n log ( n ) log ( log ( 2 ) ) k = 2 n 1 k log ( k ) log ( log ( n ) ) (2) 1 2 log ( 2 ) + 1 3 log ( 3 ) log ( log ( 3 ) )
By the Mean Value Theorem and since 1 n log ( n ) is decreasing, for some κ ( k 1 , k )
(3) 1 k log ( k ) 1 κ log ( κ ) = log ( log ( k ) ) log ( log ( k 1 ) )
Therefore, the red difference in ( 2 ) is decreasing in n since
k = 2 n 1 k log ( k ) log ( log ( n ) ) (4) = 1 2 log ( 2 ) log ( log ( 2 ) ) + k = 3 n ( 1 k log ( k ) [ log ( log ( k ) ) log ( log ( k 1 ) ) ] )
and by ( 3 ), each blue term in ( 4 ) is negative.
So, ( 2 ) says that the red difference is decreasing and bounded below by log ( log ( 2 ) ). Thus, the limit of the red difference exists and is at least log ( log ( 2 ) ) 0.366512920581664
Furthermore, ( 2 ) also says that for each n 3, the red difference is at most 1 2 log ( 2 ) + 1 3 log ( 3 ) log ( log ( 3 ) ) 0.930712768370062
Jaxon Hamilton

Jaxon Hamilton

Beginner2022-07-17Added 3 answers

The function f : x 1 x log x is continuous non negative decreasing on [ 2 , + ) hence the sequence
S n = k = 3 n ( 1 k log k k 1 k d x x log x )
is convergent, in fact
f ( n ) f ( 2 ) = k = 3 n ( f ( k ) f ( k 1 ) ) S n 0
Can you take it from here?

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