lim-x-0-x-n-1-1-n-x-1-evalute-

Question Number 196024 by York12 last updated on 16/Aug/23

\lim_{x \to 0^{+}} [x \underset{n = 1}{\sum^{\infty}} (\frac{1}{n^{x + 1}})] = λ, e v a l u t e λ

Answered by mr W last updated on 16/Aug/23

w h e n x \to 0^{+}

\frac{1}{n^{2}} < \frac{1}{n^{x + 1}} < \frac{1}{n}

\underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2}} < \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{x + 1}} < \underset{n = 1}{\sum^{\infty}} \frac{1}{n}

l e t \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{x + 1}} = A

\frac{π^{2}}{6} < A < \infty

λ = \lim_{x \to 0} (x \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{x + 1}}) = \lim_{x \to 0} (x A) = 0

Commented by York12 last updated on 17/Aug/23

t h a n k s

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