Using Riemann sums to show that sum_(i=1)^n 1/i = log(n) +c+O(1/n) I want to show that there exists a constant c such that: sum_(i=1)^n 1/i = log(n) +c+O(1/n) I am thinking about Riemann sums. Any hints on that?

Pellagra3d

Pellagra3d

Answered question

2022-10-16

Using Riemann sums to show that i = 1 n 1 / i = log n + c + O ( 1 / n )
I want to show that there exists a constant c such that:
i = 1 n 1 / i = log n + c + O ( 1 / n )
I am thinking about Riemann sums. Any hints on that?

Answer & Explanation

cokeman206

cokeman206

Beginner2022-10-17Added 18 answers

Abel's summation formula is the standard tool:
(1) n = 1 N 1 n = 1 + 1 N x x 2 d x = 1 + log N 1 N { x } x 2 d x = log N + ( 1 1 + { x } x 2 d x ) + N + { x } x 2 .
Since { x } [ 0 , 1 ), the last term is positive and bounded by N + d x x 2 = 1 N .
Paola Mayer

Paola Mayer

Beginner2022-10-18Added 1 answers

Consider integral
1 n 1 x d x = log n . Divide [ 1 , n ] to n subintervals in form [ k , k + 1 ] for k = 1 , 2 , , n 1. You have two inequalities:
1 n 1 x d x = i = 1 n 1 i i + 1 1 x d x i = 1 n 1 1 i
And
1 n 1 x d x = i = 1 n 1 i i + 1 1 x d x i = 1 n 1 1 i + 1
It's because for x [ k , k + 1 ] you have 1 k 1 x 1 k + 1

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