1-calculate-U-n-0-e-nx-4-dx-and-determine-lim-n-n-4-U-n-2-find-nature-of-the-serie-U-n-

Question Number 99578 by mathmax by abdo last updated on 21/Jun/20

1) calculate U_{n} = \int_{0}^{\infty} e^{- {nx}^{4}} dx and determine \lim_{n \to + \infty} n^{4} U_{n}

2) find nature of the serie Σ U_{n}

Answered by maths mind last updated on 22/Jun/20

U_{n}

l e t t = {n x}^{4} \Rightarrow d x = \frac{1}{n^{\frac{1}{4}}} . \frac{t^{- \frac{3}{4}}}{4} d t

= \frac{1}{4 n^{\frac{1}{4}}} \int_{0}^{+ \infty} e^{- t} t^{- \frac{3}{4}} d t = \frac{Γ (\frac{1}{4})}{4 n^{\frac{1}{4}}}

\Rightarrow n^{4} U_{n} \to + \infty

Σ U_{n} d i v e r g e Σ \frac{1}{n^{a}}, a ⩽ 1

Answered by mathmax by abdo last updated on 22/Jun/20

1) U_{n} = \int_{0}^{\infty} e^{- n x^{4}} dx we do the changement {nx}^{4} = t \Rightarrow x^{4} = \frac{t}{n} \Rightarrow

x = \frac{1}{n^{\frac{1}{4}}} t^{\frac{1}{4}} \Rightarrow U_{n} = \frac{1}{(^{4} \sqrt{n})} \int_{0}^{\infty} e^{- t} \frac{1}{4} t^{\frac{1}{4} - 1} dt = \frac{1}{4 (^{4} \sqrt{n})} \times Γ (\frac{1}{4})

U_{n} = \frac{n^{- \frac{1}{4}}}{4} Γ (\frac{1}{4}) \Rightarrow n^{4} U_{n} = \frac{n^{\frac{15}{4}}}{4} Γ (\frac{1}{4}) \to + \infty (n \to \infty)

2) \sum_{n = 1}^{\infty} U_{n} = \frac{1}{4} Γ (\frac{1}{4}) \sum_{n = 1}^{\infty} \frac{1}{n^{\frac{1}{4}}} and \sum_{n = 1}^{\infty} \frac{1}{n^{\frac{1}{4}}} have tbe same nature of

\int_{1}^{+ \infty} \frac{dt}{t^{\frac{1}{4}}} = \int_{1}^{+ \infty} t^{- \frac{1}{4}} dt = {[\frac{1}{- \frac{1}{4} + 1} t^{- \frac{1}{4} + 1}]}_{0}^{+ \infty} = {[\frac{4}{3} t^{\frac{3}{4}}]}_{0}^{\infty} = + \infty \Rightarrow

Σ u_{n} is divergent \dots