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

Question Number 38642 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^{2} + 3^{2} + \dots . + n^{2}}{1 + 2^{4} + 3^{4} + \dots . + n^{4}}

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 \dots

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

i a m n o t s o l v i n g t h e q u e s t i o n b u t p u t t i n g

s i m p l e l o g i c

2^{4} > 2^{2}

3^{4} > 3^{2}

t h u s \underset{1}{\sum^{n}} r^{4} > \underset{1}{\sum^{n}} r^{2}

s o D_{r} >> N_{r}

w h e n n \to \infty i t h i n k l i m i t v a l u e \to 0

s o p l s c o m m e n t w h e t h e r i a m r i g h t o r w r o n g

a l s o s o l v e a t l e a s t o n e q u e s t i o n t o K N O W

T H E U N K N O W N \dots

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