what is the limit as n approaches infinity of 1/n[(1/n)^9 + (2/n)^9 + (3/n)^9 + ... + (n/n)^9]

- anonymous

what is the limit as n approaches infinity of 1/n[(1/n)^9 + (2/n)^9 + (3/n)^9 + ... + (n/n)^9]

- katieb

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- myininaya

how do you evaluate sum(i^9,i=1..n) do you know?

- anonymous

no

- anonymous

please give a little more background. What class? Is this an infinite series?

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## More answers

- anonymous

is there anything else to the question

- anonymous

looks like a geometric series

- anonymous

^yeah

- anonymous

so find the ratio

- anonymous

I have to think in terms of integrals to solve it.

- anonymous

why?

- anonymous

you need a power series representation and get the ratio from that

- anonymous

the ratio is 2

- anonymous

you want
(1/n)^10 + (2/n)^10 + ...(

- anonymous

i take that back

- anonymous

the ratio of consecutive terms is not constant

- anonymous

right

- anonymous

this is an integral

- anonymous

can yuo post the original question

- anonymous

yeah thats what I asked

- anonymous

oh its a riemann sum

- anonymous

the limit is probably infinity

- anonymous

its not a decreasing function

- anonymous

no , if its a riemann sum its area under a curve

- anonymous

that means it converges

- anonymous

i just cant see the integral

- anonymous

well do you see that n is finite there

- anonymous

uphill, in that expression

- anonymous

lim (n->oo) 1/n[(1/n)^9 + (2/n)^9 + (3/n)^9 + ... + (n/n)^9]

- anonymous

n approaches infinity

- anonymous

yes but each n is finite

- anonymous

this is not a series

- anonymous

this is a rieman sum

- anonymous

maybe you need to set up an inequality

- anonymous

S sub n + the limit of the integral of that is > the limit (of stuff)

- anonymous

well the denominator is n^9

- anonymous

no you dont need that

- anonymous

lim (n->oo) 1/n[(1/n)^9 + (2/n)^9 + (3/n)^9 + ... + (n/n)^9]
= 1/n [ (1^9 + 2^9 + 3^9 + ... n^9) / n^9 ]

- anonymous

you need to find the sum formula for sum k^9 as for k = 1 to n

- anonymous

let me try and make the problem look more neat and type it again

- anonymous

give me medal

- anonymous

i solve it

- anonymous

when you pull out the /n^9 isn't it just a ratio of leading coefficients at that point? then you just have 1/n which goes to 0?

- anonymous

leading coefficients?

- anonymous

(L'pitals more or less)

- anonymous

well i dont think you can apply Lhopital;s easily

- anonymous

whichi is why u use leading coefficients

- anonymous

we need to find a formula for
1^9 + 2^9 + ... n^9

- anonymous

a closed form in terms of n

- anonymous

okay. i know formulas for powers 1,2,3 but not 9

- anonymous

and thanks you everyone for helping me out on this one. its pretty tough

- anonymous

wait, can you state the original problem again

- anonymous

sure. just a minute

- myininaya

\[\int\limits_{0}^{1}x^{9}dx=x^{10}/10, x=0..1=1/10\]

- anonymous

\[\lim_{n \rightarrow \infty} (1/n)[(1/n)^9+(2/n)^9+(3/n)^9+...+(n/n)^9]\]

- myininaya

##### 1 Attachment

- myininaya

deltax=[b-a]/n=[1-0]/n

- anonymous

right that works

- anonymous

or you can do the limit proof directly , find the sum of k^9

- anonymous

thank you very much! i finally understand it. well more than i did beforehand anyways.

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