calculate-lim-n-1-2-3-n-1-2-4-3-4-n-4-

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 + \dots + n}{1 + 2^{4} + 3^{4} + \dots + 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 + \dots + n}{1 + 2^{4} + 3^{4} + \dots + 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

\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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