calculate lim_(n→+∞) ((1+2+3+...+n)/(1+2^4 +3^4 +...+n^4 ))


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Question Number 38643 by maxmathsup by imad last updated on 27/Jun/18

$c a l c u l a t e {l i m}_{n \to + \infty} \frac{1 + 2 + 3 + . . . + n}{1 + 2^{4} + 3^{4} + . . . + n^{4}}$
Commented by abdo mathsup 649 cc last updated on 28/Jun/18

$t h e Q i s f i n d \sum_{n = 1}^{\infty} \frac{1 + 2 + 3 + . . . + n}{1 + 2^{4} + 3^{4} + . . . + n^{4}}$
Commented by tanmay.chaudhury50@gmail.com last updated on 28/Jun/18

Commented by tanmay.chaudhury50@gmail.com last updated on 28/Jun/18

Commented by tanmay.chaudhury50@gmail.com last updated on 28/Jun/18

$p l s p o s t t h e s o u r c e o f t h e q u e s t i o n a n d v a l i d i t y$
	Answered by ajfour last updated on 28/Jun/18
	$Σ_(r=1) ^n r^4 = Σ_(r=1) ^n r(r+1)(r+2)(r+3) −6Σ_(r=1) ^n r^3 −11Σ_(r=1) ^n r^2 −6Σ_(r=1) ^n r =((n(n+1)(n+2)(n+3)(n+4))/5) −((6n^2 (n+1)^2 )/4)−((11n(n+1)(2n+1))/6) −((6n(n+1))/2) = ((n(n+1))/(30)){6(n+2)(n+3)(n+4) −45n(n+1)−55(2n+1)−90} =((n(n+1))/(30)){6n^3 +9n^2 +6n−1} So lim_(n→∞) (((Σ_(r=1) ^n r)^(5/2) )/(Σ_(r=1) ^n r^4 )) = lim_(n→∞) (([((n(n+1))/2)]^(5/2) )/([((n(n+1))/(30))(6n^3 +9n^2 +6n−1)])) =((((1/2))^(5/2) )/(((1/(30)))×6)) = (5/(4(√2))) .$
	$\underset{r = 1}{\sum^{n}} r^{4} = \underset{r = 1}{\sum^{n}} r (r + 1) (r + 2) (r + 3)$ $- 6 \underset{r = 1}{\sum^{n}} r^{3} - 11 \underset{r = 1}{\sum^{n}} r^{2} - 6 \underset{r = 1}{\sum^{n}} r$ $= \frac{n (n + 1) (n + 2) (n + 3) (n + 4)}{5}$ $- \frac{6 n^{2} {(n + 1)}^{2}}{4} - \frac{11 n (n + 1) (2 n + 1)}{6}$ $- \frac{6 n (n + 1)}{2}$ $= \frac{n (n + 1)}{30} {6 (n + 2) (n + 3) (n + 4)$ $- 45 n (n + 1) - 55 (2 n + 1) - 90}$ $= \frac{n (n + 1)}{30} {6 n^{3} + 9 n^{2} + 6 n - 1}$ $S o \lim_{n \to \infty} \frac{{(\underset{r = 1}{\sum^{n}} r)}^{5 / 2}}{\underset{r = 1}{\sum^{n}} r^{4}} = \lim_{n \to \infty} \frac{{[\frac{n (n + 1)}{2}]}^{5 / 2}}{[\frac{n (n + 1)}{30} (6 n^{3} + 9 n^{2} + 6 n - 1)]}$ $= \frac{{(\frac{1}{2})}^{5 / 2}}{(\frac{1}{30}) \times 6} = \frac{5}{4 \sqrt{2}} .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 28/Jun/18

	$w h y {(\underset{r = 1}{\sum^{n}} r)}^{\frac{5}{2}}$
	Commented by ajfour last updated on 28/Jun/18

	$I d i d n o t l i k e t h e o b v i o u s a n s w e r$ $t o t h e o r i g i n a l q u e s t i o n, s o i$ $c h a n g e d t h e q u e s t i o n! (\overset{\hat{°} \hat{°}}{⌣})!$
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