Question

Use an appropriate Riemann sum to evaluate the limit
lim((1^5 + 2^5 + 3^5 + ... + n^5)/ (n^6)) as n -> infinity

Answer 1

So it would be?
\[\lim_{n \rightarrow \infty}(1^5/n^6) + \lim_{n \rightarrow \infty}(2^5/n^6) +...\lim_{n \rightarrow \infty}(n^5/n^6)\]

Answer 2

Wouldn't that equal 0?

Answer 3

@freckles

Answer 4

Yeah scratch that...Just realized something

Answer 5

Yeah I think I'm supposed to evaluate it using right endpoints, i.e. the rectangle method, but I'm not sure how to do that in a limit where n->infinity

Answer 6

Yeah scratch that...Just realized something\[\lim_{n \rightarrow \infty} \frac{1}{n^6}\sum_{i=1}^{n}i^5=\lim_{n \rightarrow \infty}\frac{1}{n^6} \cdot \frac{1}{12}n^2(n+1)^2(2n^2+2n-1)\]

Answer 7

\[\lim_{n \rightarrow \infty}\frac{2n^6 +\cdots }{12n^6 +\cdots }\]

Answer 8

you only care about the coefficient of the term with highest exponent on top and bottom

Answer 9

the degree of the top is the same as the deg on the bottom
the deg is actually 6 on top and bottom
so we only need to look at 2/12

Answer 10

How did you find that

\[\sum_{i=1}^{n}i^5 = 1/12(n^2(n+1)^2(2n^2+2n-1)\]

I remember reading that in my textbook but I can't find it again. What is this called?

Answer 11

your question says to use an appropriate riemann sum

Answer 12

that is what that is

Answer 13

I see it now. Alright, so the answer should be 1/6?

Answer 14

yes 2/12 reduces to 1/6

Answer 15

Thank you so much

Answer 16

np