solve lim_(x→∞) (ζ(x)−1)^(1/x)


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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

$\lim_{x \to \infty} {(1 + \frac{1}{2^{x}} + \frac{1}{3^{x}} + . . . . - 1)}^{\frac{1}{x}} = {(\frac{1}{2^{x}} + \frac{1}{3^{x}} + . . .)}^{x} = {(\frac{1}{2^{x}})}^{\frac{1}{x}} = \frac{1}{2}$ $\frac{1}{3^{x}} + \frac{1}{4^{x}} + . . . \to 0$
Commented by Dwaipayan Shikari last updated on 29/Sep/20

$As x \to \infty$ $\frac{1}{3^{x}} + \frac{1}{4^{x}} + . . . \to 0$ $I take only \frac{1}{2^{x}} to deal with . Others terms are so small$
	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
	$we have ξ_k (x) =Σ_(n=1) ^k (1/n^x ) ⇒ (ξ_k (x)−1)^(1/x) =(Σ_(n=2) ^k (1/n^x ))^(1/x) =e^((1/x)ln((1/2^x )+(1/3^x )+...+(1/k^x ))) =e^((1/x)ln{(1/2^x )(1+((2/3))^x +...+((2/k))^x )}) =e^((1/x){−xln2+ln(1+((2/3))^(x ) +...((2/k))^x }) =(1/2)e^((1/x)ln(1+((2/3))^x +...+((2/k))^x )) ∼(1/2) e^((1/x)(((2/3))^x +((2/4))^x +...+((2/k))^x ) ) →(1/2) for all k⇒lim_(x→+∞) lim_(k→+∞) (ξ_k (x)−1)^(1/x) =(1/2) ⇒ lim_(x→+∞) (ξ(x)−1)^(1/x) =(1/2)$
	$w e h a v e ξ_{k} (x) = \sum_{n = 1}^{k} \frac{1}{n^{x}} \Rightarrow$ ${(ξ_{k} (x) - 1)}^{\frac{1}{x}} = {(\sum_{n = 2}^{k} \frac{1}{n^{x}})}^{\frac{1}{x}}$ $= e^{\frac{1}{x} l n (\frac{1}{2^{x}} + \frac{1}{3^{x}} + . . . + \frac{1}{k^{x}})}$ $= e^{\frac{1}{x} l n {\frac{1}{2^{x}} (1 + {(\frac{2}{3})}^{x} + . . . + {(\frac{2}{k})}^{x})}}$ $= e^{\frac{1}{x} {- x l n 2 + l n (1 + {(\frac{2}{3})}^{x} + . . . {(\frac{2}{k})}^{x}}}$ $= \frac{1}{2} e^{\frac{1}{x} l n (1 + {(\frac{2}{3})}^{x} + . . . + {(\frac{2}{k})}^{x})}$ $\sim \frac{1}{2} e^{\frac{1}{x} ({(\frac{2}{3})}^{x} + {(\frac{2}{4})}^{x} + . . . + {(\frac{2}{k})}^{x})} \to \frac{1}{2}$ $f o r a l l k \Rightarrow {l i m}_{x \to + \infty} {l i m}_{k \to + \infty}$ ${(ξ_{k} (x) - 1)}^{\frac{1}{x}} = \frac{1}{2} \Rightarrow$ ${l i m}_{x \to + \infty} {(ξ (x) - 1)}^{\frac{1}{x}} = \frac{1}{2}$
	Commented by mnjuly1970 last updated on 30/Sep/20

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