let-s-put-H-n-1-2-1-3-1-n-1-and-U-n-H-n-ln-n-prove-that-U-n-is-convergent-to-a-number-s-wish-verify-0-lt-s-lt-1-s-is-named-number-of-Euler-

Question Number 25852 by abdo imad last updated on 15/Dec/17

l e t s p u t H_{n} = 1 + 2^{- 1} + 3^{- 1} + \dots . + n^{- 1} a n d U_{n} = H_{n} - l n (n)

p r o v e t h a t U_{n} i s c o n v e r g e n t t o a n u m b e r s w i s h v e r i f y

0 < s < 1 (s i s n a m e d n u m b e r o f E u l e r)

Commented by moxhix last updated on 16/Dec/17

U_{n} = \underset{k = 1}{\sum^{n}} \frac{1}{k} - \int_{1}^{n} \frac{1}{x} d x

= \underset{k = 1}{\sum^{n}} \frac{1}{k} - \underset{k = 1}{\sum^{n - 1}} \int_{k}^{k + 1} \frac{1}{x} d x

= \frac{1}{n} + \underset{k = 1}{\sum^{n - 1}} (\frac{1}{k} - \int_{k}^{k + 1} \frac{1}{x} d x)

= \frac{1}{n} + \underset{k = 1}{\sum^{n - 1}} (\int_{k}^{k + 1} \frac{1}{k} - \frac{1}{x} d x)

> (\frac{1}{1} - \int_{1}^{2} \frac{1}{x} d x)

= 1 - l n (2)

∴ U_{n} > 1 - l n (2) (\forall n ⩾ 2)

U_{n} = \underset{k = 1}{\sum^{n}} \frac{1}{k} - \int_{1}^{n} \frac{1}{x} d x

= \underset{k = 1}{\sum^{n}} \frac{1}{k} - \underset{k = 2}{\sum^{n}} \int_{k - 1}^{k} \frac{1}{x} d x

= 1 + \underset{k = 2}{\sum^{n}} (\frac{1}{k} - \int_{k - 1}^{k} \frac{1}{x} d x)

= 1 + \underset{k = 2}{\sum^{n}} (\int_{k - 1}^{k} \frac{1}{k} - \frac{1}{x} d x)

< 1 + \int_{1}^{2} \frac{1}{2} - \frac{1}{x} d x

= \frac{3}{2} - l n (2)

∴ U_{n} < \frac{3}{2} - l n (2) (\forall n ⩾ 2)

U_{n + 1} - U_{n} = \frac{1}{n + 1} - \int_{n}^{n + 1} \frac{1}{x} d x

= \int_{n}^{n + 1} \frac{1}{n + 1} - \frac{1}{x} d x

< 0

∴ U_{n + 1} < U_{n}

↓

1 - l n (2) < \dots < U_{n + 1} < U_{n} < \dots < U_{3} < U_{2} < \frac{3}{2} - l n (2)

Double subscripts: use braces to clarify

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