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Question Number 58354 by maxmathsup by imad last updated on 21/Apr/19

l e t U_{n} = \frac{1^{2} + 2^{2} + 3^{2} + \dots . + n^{2}}{1^{4} + 2^{4} + 3^{4} + \dots . + n^{4}}

1) f i n d {l i m}_{n \to + \infty} U_{n}

2) c a l c u l a t e \sum_{n = 1}^{\infty} U_{n}

Answered by tanmay last updated on 22/Apr/19

U_{n} h e r e = \frac{N_{r}}{D_{r}}

N_{r} = \underset{r = 1}{\sum^{n}} r^{2} = \frac{n^{3}}{3} + \frac{n^{2}}{2} + \frac{n}{6}

D_{r} = \underset{r = 1}{\sum^{n}} r^{4} = \frac{n^{5}}{5} + \frac{n^{4}}{2} + \frac{n^{3}}{3} - \frac{n}{30}

e a c h t e r m o f N_{r} < e a c h t e r m o f D_{r}

\lim_{n \to \infty} \frac{\frac{n^{3}}{3} + \frac{n^{2}}{2} + \frac{n}{6}}{\frac{n^{5}}{5} + \frac{n^{4}}{2} + \frac{n^{3}}{3} - \frac{n}{30}}

d e v i d e N_{r} a n d D_{r} b y n^{5} \to

\lim_{n \to \infty} \frac{\frac{1}{3 n^{2}} + \frac{1}{2 n^{3}} + \frac{1}{6 n^{4}}}{\frac{1}{5} + \frac{1}{2 n} + \frac{1}{3 n^{2}} - \frac{1}{30 n^{4}}}

= \frac{0 + 0 + 0}{\frac{1}{5} + 0 + 0 - 0}

= 0

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