Question

Find the value of:
\[\large \lim_{n \rightarrow \infty } \left(\frac{1^{3}+2^{3}+...+n^{3}}{n^{4}}\right)\]

Answer 1

\[e^{i \theta} = \cos(\theta)+isin(\theta)\]
Can you work it out using this?

Answer 2

Chances are you can.

Answer 3

Yes you can, I'm trying to encourage him to work out the answer himself.

Answer 4

\[\lim_{x \rightarrow infinity} \left(\frac{1^{3}+2^{3}+...+n^{3}}{n^{4}}\right) \]

Answer 5

lol, sorry

Answer 6

The answer isn't 0 lol.

Answer 7

pardon?

Answer 8

i never said that its 0..

Answer 9

If that x under the limit was an n (which I assumed it was meant to be) then it would tend to 0.

Answer 10

Anyway back to the question, are you able to work it out?

Answer 11

i mean n tend to infinity

Answer 12

Then that limit tends to 0, which isn't the right answer.

Answer 13

its \[\large \lim_{n \rightarrow \infty } \left(\frac{1^{3}+2^{3}+...+n^{3}}{n^{4}}\right)\]

Answer 14

=0

Answer 15

its not 0..

Answer 16

how can it be 0 tho

Answer 17

You have:
\[\lim_{n \rightarrow \infty}(\frac{ \sum_{k=1}^{n}k^3 }{ k^4 })=\lim_{n \rightarrow \infty}( \sum_{k=1}^{n}\frac{ 1 }{ k } )=0\]

Answer 18

Or to see it easier, if you divide each term on the top and bottom of the fraction by n^4 you will get lots of fractions all of which tend to 0 as n->infinity

Answer 19

woops its\[\large \lim_{n \rightarrow \infty } \left(\frac{1^{3}+2^{3}+ 3^{3} ...+n^{3}}{n^{4}}\right)\]
but i dont think its 0..

Answer 20

its a Riemann sum change it to a simple integral

Answer 21

Wait no I'm not, I just didn't type it in properly.

Answer 22

the answer is 1/4

Answer 23

hm..how do i change it into a simple integral?

Answer 24

The top of the fraction, sum of k^3 from 1 to n, is 1/4 n^2(n+1)^2. So after cancelling you will get 1/4. Anyway, 1/4 is still not the answer to the original question so I don't see why we're talking about this...

Answer 25

what math u are in Omn?

Answer 26

is this from calculus?

Answer 27

well..yes

Answer 28

\[\large \lim_{n \rightarrow \infty } \left(\frac{1^{3}+2^{3}+ 3^{3} ...+n^{3}}{n^{4}}\right)=\lim_{n \rightarrow \infty } \left((\frac{1}{n})^3+(\frac{2}{n})^3+...+(\frac{n}{n})^3\right)\frac{1}{n}\\ \large=\int_{0}^{1}x^3 \text{d}x\]

Answer 29

otherwise use the traxter method

Answer 30

oohh..i got it thanks.

Answer 31

:)