solve-lim-x-x-1-1-x-

Question Number 115812 by mathdave last updated on 28/Sep/20

s o l v e

\lim_{x \to \infty} {(ζ (x) - 1)}^{\frac{1}{x}}

Commented by Dwaipayan Shikari last updated on 28/Sep/20

Is it \lim_{x \to \infty} {(ζ (x) - 1)}^{\frac{1}{x}} ? ?

Commented by mathdave last updated on 28/Sep/20

y e s a n y i d e a

Commented by Dwaipayan Shikari last updated on 28/Sep/20

\lim_{x \to \infty} {(1 + \frac{1}{2^{x}} + \frac{1}{3^{x}} + \dots . - 1)}^{\frac{1}{x}} = {(\frac{1}{2^{x}} + \frac{1}{3^{x}} + \dots)}^{x} = {(\frac{1}{2^{x}})}^{\frac{1}{x}} = \frac{1}{2}

\frac{1}{3^{x}} + \frac{1}{4^{x}} + \dots \to 0

Commented by Dwaipayan Shikari last updated on 29/Sep/20

As x \to \infty

\frac{1}{3^{x}} + \frac{1}{4^{x}} + \dots \to 0

I take only \frac{1}{2^{x}} to deal with . Others terms are so small

Answered by Bird last updated on 29/Sep/20

i f o u n d \frac{1}{2} i s i t c o r r e c t ?

Commented by mathdave last updated on 29/Sep/20

i t i s c o r r e c t

Answered by Bird last updated on 29/Sep/20

w e h a v e ξ_{k} (x) = \sum_{n = 1}^{k} \frac{1}{n^{x}} \Rightarrow

{(ξ_{k} (x) - 1)}^{\frac{1}{x}} = {(\sum_{n = 2}^{k} \frac{1}{n^{x}})}^{\frac{1}{x}}

= e^{\frac{1}{x} l n (\frac{1}{2^{x}} + \frac{1}{3^{x}} + \dots + \frac{1}{k^{x}})}

= e^{\frac{1}{x} l n {\frac{1}{2^{x}} (1 + {(\frac{2}{3})}^{x} + \dots + {(\frac{2}{k})}^{x})}}

= e^{\frac{1}{x} {- x l n 2 + l n (1 + {(\frac{2}{3})}^{x} + \dots {(\frac{2}{k})}^{x}}}

= \frac{1}{2} e^{\frac{1}{x} l n (1 + {(\frac{2}{3})}^{x} + \dots + {(\frac{2}{k})}^{x})}

\sim \frac{1}{2} e^{\frac{1}{x} ({(\frac{2}{3})}^{x} + {(\frac{2}{4})}^{x} + \dots + {(\frac{2}{k})}^{x})} \to \frac{1}{2}

f o r a l l k \Rightarrow {l i m}_{x \to + \infty} {l i m}_{k \to + \infty}

{(ξ_{k} (x) - 1)}^{\frac{1}{x}} = \frac{1}{2} \Rightarrow

{l i m}_{x \to + \infty} {(ξ (x) - 1)}^{\frac{1}{x}} = \frac{1}{2}

Commented by mnjuly1970 last updated on 30/Sep/20

n i c e v e r y n i c e