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Question Number 65945 by Chi Mes Try last updated on 06/Aug/19

p l s i n e e d s o l u t i o n p l s s s s \dots a s a p

n

l i m \in (\frac{r^{3}}{r^{4} + n^{4}})

n \to \infty r = 1

p l e a s e t r y a n d u n d e r s t a n d t h e w a y i t y p e d i t

Commented by MJS last updated on 06/Aug/19

\lim_{n \to \infty} \underset{r = 1}{\sum^{n}} \frac{r^{3}}{r^{4} + n^{4}}

is this what you mean ?

Commented by mathmax by abdo last updated on 06/Aug/19

l e t S_{n} = \sum_{k = 1}^{n} \frac{k^{3}}{n^{4} + k^{4}} \Rightarrow S_{n} = \sum_{k = 1}^{n} \frac{1}{n} \times \frac{{(\frac{k}{n})}^{3}}{1 + {(\frac{k}{n})}^{4}}

s o S_{n} i s R i e m a n s u m a n d {l i m}_{n \to + \infty} S_{n} = \int_{0}^{1} \frac{x^{3}}{1 + x^{4}} d x

= {[\frac{1}{4} l n (1 + x^{4})]}_{0}^{1} = \frac{l n (2)}{4}