Compute Power Series Convergence to a function Consider the next power series sum_(n=1)^(oo) ln (n) z^n Find the convergence radius and a the function f to which the series converges. I have easily found that R=1 is the convergence radius, however I can not find the function. I was trying to found an elemental function with this power series expantion, but I have failed. Anyone knows such function and how to prove the convergence?

George Morales 2022-10-13 Answered
Compute Power Series Convergence to a function
Consider the next power series
n = 1 ln ( n ) z n
Find the convergence radius and a the function f to which the series converges.
I have easily found that R = 1 is the convergence radius, however I can not find the function. I was trying to found an elemental function with this power series expantion, but I have failed. Anyone knows such function and how to prove the convergence?
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Answers (2)

blogpolisft
Answered 2022-10-14 Author has 10 answers
Here is an approach.
Assume | z | < 1
From 0 ln n < n, n = 1 , 2 , 3 , , we get
| n = 1 ln n z n | n = 1 ln n | z | n n = 1 n | z | n = | z | ( 1 | z | ) 2 <
Assume | z | 1
Then
lim z | ln n z n | = 0.
This proves that our power series admits a radius of convergence equal to 1.
Let z be a complex number such that | z | < 1, then
(1) n = 1 ln n z n = z log ( 1 z ) z 2 γ ( z ) 1 z
where γ ( ) denotes a special function called the generalized-Euler-constant function which has been studied by Jonathan Sondow and Petros Hadjicostas, amongst others.
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Iris Vaughn
Answered 2022-10-15 Author has 3 answers
Here is another answer. Clearly, the convergence radius is given by
R = lim n ln ( n ) ln ( n + 1 ) = lim n n + 1 n = 1
Consider now the polylogaritmic function given by
Li s ( z ) := n = 1 z n n s .
Then is easily seen that for z { z C : | z | < 1 } we have
n = 1 ln ( n ) z n = ( s Li s ( z ) ) s = 0
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