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Question Number 47725 by Meritguide1234 last updated on 14/Nov/18

Commented by Meritguide1234 last updated on 15/Nov/18

$a n s i s \sqrt{2} + 1$
Commented by maxmathsup by imad last updated on 15/Nov/18

$t h i s a n s w e r i s n o t c o r r e c t s i r w h e r e c o m e t h e n u m b e r \sqrt{2} ? . . . .$
	Answered by maxmathsup by imad last updated on 15/Nov/18
	$if the Q. is find lim_(n→+∞) ((Σ_(k=0) ^n (√((√n)+(√k))))/(Σ_(k=0) ^n (√((√n)−(√k))))) let S_n =((Σ_(k=0) ^n (√((√n)+(√k))))/(Σ_(k=0) ^n (√((√n)−(√k))))) we have S_n =((n^(1/4) Σ_(k=0) ^n (√(1+(√(k/n)))))/(n^(1/4) (√(1−(√(k/n)))))) =(((1/n)Σ_(k=0) ^n (√(1+(√(k/n)))))/((1/(n ))Σ_(k=0) ^n (√(1−(√(k/n)))))) so S_n is a Rieman sum and lim_(n→+∞) S_n =((∫_0 ^1 (√(1+(√x)))dx)/(∫_0 ^1 (√(1−(√x)))dx)) let calculate ∫_0 ^1 (√(1+(√x)))dx changement (√(1+(√x)))=t give 1+(√x)=t^2 ⇒(√x)=t^2 −1 ⇒x=(t^2 −1)^2 =t^4 −2t^2 +1 ⇒ dx=4t^3 −4t)dt ⇒∫_0 ^1 (√(1+(√x)))dx =∫_1 ^2 t(4t^3 −4t)dt =4 ∫_1 ^2 (t^4 −t^2 )dt =4[(t^5 /5) −(t^3 /3)]_1 ^2 =4 {(2^5 /5) −(2^3 /3) } =32{ (4/5) −(1/3)} =32.(7/(15)) =((224)/(15)) now let calculate ∫_0 ^1 (√(1−(√x)))dx changement (√(1−(√x)))=t give 1−(√x)=t^2 ⇒(√x)=1−t^2 ⇒ x=(1−t^2 )^2 =t^4 −2t^2 +1 ⇒dx =(4t^3 −4t)dt ⇒ ∫_0 ^1 (√(1−(√x)))dx = −∫_0 ^1 t(4t^3 −4t)dt =−4 ∫_0 ^1 (t^4 −t^2 )dt =4 ∫_0 ^1 (t^2 −t^4 )dt =4[(t^3 /3) −(t^5 /5)]_0 ^1 =4{(1/3) −(1/5)}=4.(2/(15)) =(8/(15)) ⇒ lim_(n→+∞) S_n =((224)/8) =((112)/4) =((56)/2) =28 .$
	$i f t h e Q . i s f i n d {l i m}_{n \to + \infty} \frac{\sum_{k = 0}^{n} \sqrt{\sqrt{n} + \sqrt{k}}}{\sum_{k = 0}^{n} \sqrt{\sqrt{n} - \sqrt{k}}} l e t S_{n} = \frac{\sum_{k = 0}^{n} \sqrt{\sqrt{n} + \sqrt{k}}}{\sum_{k = 0}^{n} \sqrt{\sqrt{n} - \sqrt{k}}}$ $w e h a v e S_{n} = \frac{n^{\frac{1}{4}} \sum_{k = 0}^{n} \sqrt{1 + \sqrt{\frac{k}{n}}}}{n^{\frac{1}{4}} \sqrt{1 - \sqrt{\frac{k}{n}}}} = \frac{\frac{1}{n} \sum_{k = 0}^{n} \sqrt{1 + \sqrt{\frac{k}{n}}}}{\frac{1}{n} \sum_{k = 0}^{n} \sqrt{1 - \sqrt{\frac{k}{n}}}} s o S_{n} i s a R i e m a n s u m$ $a n d {l i m}_{n \to + \infty} S_{n} = \frac{\int_{0}^{1} \sqrt{1 + \sqrt{x}} d x}{\int_{0}^{1} \sqrt{1 - \sqrt{x}} d x} l e t c a l c u l a t e \int_{0}^{1} \sqrt{1 + \sqrt{x}} d x c h a n g e m e n t$ $\sqrt{1 + \sqrt{x}} = t g i v e 1 + \sqrt{x} = t^{2} \Rightarrow \sqrt{x} = t^{2} - 1 \Rightarrow x = {(t^{2} - 1)}^{2} = t^{4} - 2 t^{2} + 1 \Rightarrow$ $d x = 4 t^{3} - 4 t) d t \Rightarrow \int_{0}^{1} \sqrt{1 + \sqrt{x}} d x = \int_{1}^{2} t (4 t^{3} - 4 t) d t = 4 \int_{1}^{2} (t^{4} - t^{2}) d t$ $= 4 {[\frac{t^{5}}{5} - \frac{t^{3}}{3}]}_{1}^{2} = 4 {\frac{2^{5}}{5} - \frac{2^{3}}{3}} = 32 {\frac{4}{5} - \frac{1}{3}} = 32 . \frac{7}{15} = \frac{224}{15} n o w l e t c a l c u l a t e$ $\int_{0}^{1} \sqrt{1 - \sqrt{x}} d x c h a n g e m e n t \sqrt{1 - \sqrt{x}} = t g i v e 1 - \sqrt{x} = t^{2} \Rightarrow \sqrt{x} = 1 - t^{2} \Rightarrow$ $x = {(1 - t^{2})}^{2} = t^{4} - 2 t^{2} + 1 \Rightarrow d x = (4 t^{3} - 4 t) d t \Rightarrow$ $\int_{0}^{1} \sqrt{1 - \sqrt{x}} d x = - \int_{0}^{1} t (4 t^{3} - 4 t) d t = - 4 \int_{0}^{1} (t^{4} - t^{2}) d t$ $= 4 \int_{0}^{1} (t^{2} - t^{4}) d t = 4 {[\frac{t^{3}}{3} - \frac{t^{5}}{5}]}_{0}^{1} = 4 {\frac{1}{3} - \frac{1}{5}} = 4 . \frac{2}{15} = \frac{8}{15} \Rightarrow$ ${l i m}_{n \to + \infty} S_{n} = \frac{224}{8} = \frac{112}{4} = \frac{56}{2} = 28 .$
	Commented by Meritguide1234 last updated on 16/Nov/18

	Commented by Meritguide1234 last updated on 16/Nov/18

	$i t g i v e s s a m e r e s u l t i f y o u o p t e d t o t a k e l i m i t a n d c o n v e r t i n l i m i t f o r m$
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