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Question Number 27785 by abdo imad last updated on 14/Jan/18

f i n d n a t u r e o f t h e s e r i e \sum_{n = 1}^{\propto} ξ (n) x^{n}

w i t h ξ (x) = \sum_{n = 1}^{\propto} \frac{1}{n^{x}} a n d x > 1 .

Commented by abdo imad last updated on 20/Jan/18

l e t p u t S (x) = \sum_{n = 1}^{\infty} ξ (n) x^{n} i f x = 0 S (x) = 0 l e t s u p p o s e

x \neq 0 w e h a v e ∣ \frac{ξ (n + 1) x^{n + 1}}{ξ (n) x^{n}} ∣= \frac{ξ (n + 1)}{ξ (n)} ∣ x ∣ l e t s t u d i e

{l i m}_{n \to + \infty^{}} \frac{ξ (n + 1)}{ξ (n)} w e h a v e

ξ (n + 1) = \sum_{p = 1}^{\propto} \frac{1}{p^{n + 1}} a n d ξ (n) = \sum_{p = 1}^{\propto} \frac{1}{p^{n}}

\frac{1}{p^{n + 1}} = \frac{1}{p . p^{n}} < \frac{1}{p^{n}} f o r p > 1 \Rightarrow 0 < ξ (n + 1) < ξ (n)

\Rightarrow 0 < \frac{ξ (n + 1)}{ξ (n)} < 1 i f l = {l i m}_{n \to + \propto} \frac{ξ (n + 1)}{ξ (n)} w e m u s t h a v e

0 < l < 1 a n d {l i m}_{n \to + \propto} \frac{ξ (n + 1)}{ξ (n)} / x / = l / x / i f

/ x / < \frac{1}{l} t h e s e r i e c n v e r g e s a n d i f / x / > \frac{1}{l} t h e s e r i e

d i v e r g e s .