advanced-calculus-prove-that-lim-n-1-n-1-n-n-1-n-d-n-dx-n-ln-x-x-x-n-euler-mascheroni-constant-

Question Number 139530 by mnjuly1970 last updated on 28/Apr/21

\dots \dots . a d v a n c e d c a l c u l u s \dots \dots

p r o v e t h a t ::

{l i m}_{n \to \infty} {\frac{{(- 1)}^{n + 1} n^{n + 1}}{n!} \frac{d^{n}}{{d x}^{n}} (\frac{l n (x)}{x}) ∣_{x = n}} = γ

γ : e u l e r - m a s c h e r o n i c o n s t a n t

Answered by mindispower last updated on 28/Apr/21

\frac{d^{n}}{{d x}^{n}} \frac{l n (x)}{x} = \underset{k = 0}{\sum^{n}} C_{n}^{k} {(l n (x))}^{k} . {(\frac{1}{x})}^{n - k} . L i b n e i z f o r m u l a

l n {(x)}^{(k)} = l n (x), k = 0

= {(\frac{1}{x})}^{(k - 1)}, k ⩾ 1

= \frac{{(- 1)}^{k - 1} . (k - 1)!}{x^{k - 1}}, k ⩾ 1

{(\frac{1}{x})}^{n - k} = \frac{{(- 1)}^{n - k} (n - k)!}{x^{n - k + 1}}

= (\underset{k = 1}{\sum^{n}} C_{n}^{k} \frac{{(- 1)}^{n - 1} (k - 1)! . (n - k)!}{x^{n + 1}} + l n (x) . \frac{{(- 1)}^{n} n!}{x^{n + 1}}) ∣_{x = n} . \frac{{(- 1)}^{n + 1} n^{n + 1}}{n!}

= \sum_{k ⩾ 1} \frac{{(- 1)}^{n - 1}}{n^{n + 1}} . \frac{(k - 1)! . (n - k)!}{n^{n + 1}} . \frac{n!}{k! . (n - k)!} . \frac{{(- 1)}^{n + 1}}{n!} n^{n + 1}

+ l n (n) . \frac{{(- 1)}^{n}}{n^{n + 1}} n! . \frac{{(- 1)}^{n + 1} n^{n + 1}}{n!}

= \underset{k = 1}{\sum^{n}} \frac{1}{k} - l n (n)

\lim_{n \to \infty} \underset{k = 1}{\sum^{n}} \frac{1}{k} - l n (n) = γ B y d e f i n i t i o n

Commented by mnjuly1970 last updated on 28/Apr/21

t h a n k s a l o t s i r p o w e r \dots

m e r c e y \dots .

Commented by mindispower last updated on 29/Apr/21

p l e a s u r