prove-that-u-n-k-1-n-1-n-k-is-convergente-

Question Number 30175 by abdo imad last updated on 17/Feb/18

p r o v e t h a t u_{n} = \sum_{k = 1}^{n} \frac{1}{n + k} i s c o n v e r g e n t e .

Commented by abdo imad last updated on 21/Feb/18

w e h a v e u_{n} = \frac{1}{n} \sum_{k = 1}^{n} \frac{1}{1 + \frac{k}{n}} \Rightarrow

{l i m}_{n \to \infty} u_{n} = {l i m}_{n \to \infty} \frac{1 - 0}{n} \sum_{k = 1}^{n} \frac{1}{1 + \frac{k (1 - 0)}{n}} (R i e m a n s u m)

= \int_{0}^{1} \frac{d x}{1 + x} = {[l n ∣ 1 + x ∣]}_{0}^{1} = l n 2 .

Commented by abdo imad last updated on 21/Feb/18

a n o t h e r m e t h o d w e h a v e

u_{n} = \frac{1}{n + 1} + \frac{1}{n + 2} + \dots . \frac{1}{2 n} = 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n} + \frac{1}{n + 1} + \dots

+ \frac{1}{2 n} - (1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}) = H_{2 n} - H_{n} b u t

H_{2 n} = l n (2 n) + γ + o (\frac{1}{n}) a n d H_{n} = l n (n) + γ + o (\frac{1}{n}) \Rightarrow

H_{2 n} - H_{n} = l n (\frac{2 n}{n}) + o (\frac{1}{n}) \Rightarrow {l i m}_{n \to \infty} H_{2 n} - H_{n} = l n (2) . s o

{l i m}_{n \to \infty} u_{n} = l n (2) .