let ξ(x) =Σ_(n=1) ^∞ (1/n^x ) calculate lim


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Question Number 98189 by abdomathmax last updated on 12/Jun/20

$let ξ (x) = \sum_{n = 1}^{\infty} \frac{1}{n^{x}}$ $calculate \lim_{x \to 1^{+}} (x - 1) ξ (x)$
	Answered by mathmax by abdo last updated on 12/Jun/20
	$the function t→(1/t^x ) is decreazing on ]1,+∞[ so Σ_(k=2) ^n ∫_k ^(k+1) (dt/t^x ) ≤ Σ_(k=2) ^n (1/k^x ) ≤Σ_(k=2) ^n ∫_(k−1) ^k (dt/t^x ) ⇒ ∫_2 ^(n+1) (dt/t^x ) ≤ ξ_n (x) ≤∫_1 ^n (dt/t^x ) we have ∫_2 ^(n+1) (dt/t^x ) =∫_2 ^(n+1) t^(−x) dt =[(1/(1−x))t^(−x+1) ]_2 ^(n+1) =(1/(1−x)){ (1/((n+1)^(x−1) ))−(1/2^(x−1) )} =(1/((x−1))){(1/2^(x−1) )−(1/((n+1)^(x−1) ))} ∫_1 ^n t^(−x) dt =[(1/(−x+1)) t^(−x+1) ]_1 ^n =(1/(1−x))[ (1/t^(x−1) )]_1 ^n =(1/(1−x)){(1/n^(x−1) )−1} =(1/(x−1)){1−(1/n^(x−1) )} ⇒ (1/(x−1)){(1/2^(x−1) )−(1/((n+1)^(x−1) ))} ≤ Σ_(k=1) ^n (1/k^x )−1 ≤(1/(x−1)){1−(1/n^(x−1) )} ⇒ ∀x>1 (1/2^(x−1) ) −(1/((n+1)^(x−1) )) ≤(x−1)ξ_n (x)−(x−1)≤1−(1/n^(x−1) ) ⇒ ⇒(1/2^(x−1) ) ≤ (x−1)ξ(x)−(x−1)≤ 1 ⇒lim_(x→1^+ ) (1/2^(x−1) ) ≤lim_(x→1^+ ) (x−1)ξ(x)≤1 ⇒lim_(x→1^+ ) (x−1)ξ(x) =1$
	$the function t \to \frac{1}{t^{x}} is decreazing on] 1, + \infty [so$ $\sum_{k = 2}^{n} \int_{k}^{k + 1} \frac{dt}{t^{x}} ⩽ \sum_{k = 2}^{n} \frac{1}{k^{x}} ⩽ \sum_{k = 2}^{n} \int_{k - 1}^{k} \frac{dt}{t^{x}} \Rightarrow$ $\int_{2}^{n + 1} \frac{dt}{t^{x}} ⩽ ξ_{n} (x) ⩽ \int_{1}^{n} \frac{dt}{t^{x}} we have \int_{2}^{n + 1} \frac{dt}{t^{x}} = \int_{2}^{n + 1} t^{- x} dt = {[\frac{1}{1 - x} t^{- x + 1}]}_{2}^{n + 1}$ $= \frac{1}{1 - x} {\frac{1}{{(n + 1)}^{x - 1}} - \frac{1}{2^{x - 1}}} = \frac{1}{(x - 1)} {\frac{1}{2^{x - 1}} - \frac{1}{{(n + 1)}^{x - 1}}}$ $\int_{1}^{n} t^{- x} dt = {[\frac{1}{- x + 1} t^{- x + 1}]}_{1}^{n} = \frac{1}{1 - x} {[\frac{1}{t^{x - 1}}]}_{1}^{n} = \frac{1}{1 - x} {\frac{1}{n^{x - 1}} - 1}$ $= \frac{1}{x - 1} {1 - \frac{1}{n^{x - 1}}} \Rightarrow \frac{1}{x - 1} {\frac{1}{2^{x - 1}} - \frac{1}{{(n + 1)}^{x - 1}}} ⩽ \sum_{k = 1}^{n} \frac{1}{k^{x}} - 1 ⩽ \frac{1}{x - 1} {1 - \frac{1}{n^{x - 1}}} \Rightarrow$ $\forall x > 1 \frac{1}{2^{x - 1}} - \frac{1}{{(n + 1)}^{x - 1}} ⩽ (x - 1) ξ_{n} (x) - (x - 1) ⩽ 1 - \frac{1}{n^{x - 1}} \Rightarrow$ $\Rightarrow \frac{1}{2^{x - 1}} ⩽ (x - 1) ξ (x) - (x - 1) ⩽ 1 \Rightarrow \lim_{x \to 1^{+}} \frac{1}{2^{x - 1}} ⩽ \lim_{x \to 1^{+}} (x - 1) ξ (x) ⩽ 1$ $\Rightarrow \lim_{x \to 1^{+}} (x - 1) ξ (x) = 1$
	Commented by mathmax by abdo last updated on 12/Jun/20

	$the important result here is that if f decrease we have$ $\int_{k}^{k + 1} f (t) dt ⩽ f (k) ⩽ \int_{k - 1}^{k} f (t) dt here f (t) = \frac{1}{t^{x}} (t > 1)$
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