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


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Question Number 38642 by maxmathsup by imad last updated on 27/Jun/18

$c a l c u l a t e {l i m}_{n \to + \infty} \frac{1 + 2^{2} + 3^{2} + . . . . + n^{2}}{1 + 2^{4} + 3^{4} + . . . . + n^{4}}$
Commented by tanmay.chaudhury50@gmail.com last updated on 28/Jun/18

$p l s p o s t t h e s o u r c e o f t h e q u e s t i o n . . .$
Commented by tanmay.chaudhury50@gmail.com last updated on 28/Jun/18

$i a m n o t s o l v i n g t h e q u e s t i o n b u t p u t t i n g$ $s i m p l e l o g i c$ $2^{4} > 2^{2}$ $3^{4} > 3^{2}$ $t h u s \underset{1}{\sum^{n}} r^{4} > \underset{1}{\sum^{n}} r^{2}$ $s o D_{r} >> N_{r}$ $w h e n n \to \infty i t h i n k l i m i t v a l u e \to 0$ $s o p l s c o m m e n t w h e t h e r i a m r i g h t o r w r o n g$ $a l s o s o l v e a t l e a s t o n e q u e s t i o n t o K N O W$ $T H E U N K N O W N . . .$
Commented by abdo mathsup 649 cc last updated on 28/Jun/18

$t h e Q i s f i n d \sum_{n = 1}^{\infty} \frac{1 + 2^{2} + 3^{2} + . . . + n^{2}}{1 + 2^{4} + 3^{4} + . . . + n^{4}}$
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