Limit of a harmonic subseries minus "its" logarithm lim n → ∞ ∑ k =...

Eden Solomon

Eden Solomon

Answered

2022-06-26

Limit of a harmonic subseries minus "its" logarithm
lim n k = 1 n 1 3 k 1 1 3 ln ( n )
I think that inserting the other terms and then subtracting them would not help. I need just the ideea. Thank you.

Answer & Explanation

Quinn Everett

Quinn Everett

Expert

2022-06-27Added 23 answers

It is not very difficult to show that, for natural values of a, k = 1 n 1 n + a = H n + a H a , where H n is the n t h harmonic number. But in our case, a = 1 3 , so we have to extend the formula or definition of H n so as to encompass non-natural arguments as well. Luckily for us, Euler has already done it some 250 years ago: H n = 0 1 1 x n 1 x d x . Using this, we can now compute the value of H 1 3 to be 1 2 ( π 3 3 ln 3 ) . Now use the fact that lim n H n ln n = γ . See here for more details on how to handle the integral in question.
pokoljitef2

pokoljitef2

Expert

2022-06-28Added 9 answers

If it were k = 1 n 1 k ln n , we'd end up with γ. So I'd suggest to investigate
k = 1 ( 3 3 k 1 1 k )

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