I have come across the following series, which I suspect converges to ln2: sum_(k=1)^oo (1)(4^k(2k))((2k),(k)) I could not derive this series from some of the standard expressions for ln2. The sum of the first 100000 terms agrees with ln2 only up to two digits.

Macioccujx 2022-07-16 Answered
A series converging (or not) to ln 2
I have come across the following series, which I suspect converges to ln 2
k = 1 1 4 k ( 2 k ) ( 2 k k ) .
I could not derive this series from some of the standard expressions for ln 2. The sum of the first
100 000 terms agrees with ln 2 only up to two digits.
Does the series converge to ln 2?
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Answers (2)

Abbigail Vaughn
Answered 2022-07-17 Author has 15 answers
Hint: (or outline -- a lot of details and justifications must be done where there are ( )'s)
for | x | < 1 4 ,
f ( x ) = d e f 1 2 k = 1 ( 2 k k ) x k k = 1 2 k = 1 ( 2 k k ) 0 x t k 1 d t = ( ) 1 2 0 x ( k = 1 ( 2 k k ) t k 1 ) d t = 1 2 0 x 1 t ( k = 1 ( 2 k k ) t k ) d t = ( ) 1 2 0 x 1 t ( 1 1 4 t 1 ) d t = 1 2 2 ln 2 1 4 x + 1
"so" ( )
( ) f ( x ) t 1 4 ln 2
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ganolrifv9
Answered 2022-07-18 Author has 4 answers
Recall that for | x | < 1 we have
1 1 x 2 = k = 0 ( 2 n n ) x 2 n 4 n
shifting over the first terms on the series and dividing by x gives
n = 1 ( 2 n n ) x 2 n 1 4 n = 1 1 x 2 1 x 2 1 x
Now given an small ε > 0 we can integrate both sides from 0 to 1 ε, the swap on the RHS is justified due to the uniform convergence, next we let ε shrink to zero. Wallis product formula yields
1 4 n 2 n ( 2 n n ) 1 2 π 1 n 2 n + 1
Now given an small ε > 0 we can integrate both sides from 0 to 1 ε, the swap on the RHS is justified due to the uniform convergence, next we let ε shrink to zero. Wallis product formula yields
1 4 n 2 n ( 2 n n ) 1 2 π 1 n 2 n + 1
which allows us to move in the limit inside and letting ε 0 Hence we have that
n = 1 ( 2 n n ) 1 4 n 2 n = 0 1 1 1 x 2 1 x 2 d x x = 0 π / 2 1 cos ( u ) sin ( u ) d u = 0 π / 2 sin ( u / 2 ) cos ( u / 2 ) d u
It is pretty obvious what to do next..
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