let-x-n-1-1-n-x-calculate-lim-x-1-x-1-x-

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 \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)