Question

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

Answer 1

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

Answer 2

no

Answer 3

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

Answer 4

is there anything else to the question

Answer 5

looks like a geometric series

Answer 6

^yeah

Answer 7

so find the ratio

Answer 8

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

Answer 9

why?

Answer 10

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

Answer 11

the ratio is 2

Answer 12

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

Answer 13

i take that back

Answer 14

the ratio of consecutive terms is not constant

Answer 15

right

Answer 16

this is an integral

Answer 17

can yuo post the original question

Answer 18

yeah thats what I asked

Answer 19

oh its a riemann sum

Answer 20

the limit is probably infinity

Answer 21

its not a decreasing function

Answer 22

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

Answer 23

that means it converges

Answer 24

i just cant see the integral

Answer 25

well do you see that n is finite there

Answer 26

uphill, in that expression

Answer 27

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

Answer 28

n approaches infinity

Answer 29

yes but each n is finite

Answer 30

this is not a series

Answer 31

this is a rieman sum

Answer 32

maybe you need to set up an inequality

Answer 33

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

Answer 34

well the denominator is n^9

Answer 35

no you dont need that

Answer 36

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 ]

Answer 37

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

Answer 38

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

Answer 39

give me medal

Answer 40

i solve it

Answer 41

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?

Answer 42

leading coefficients?

Answer 43

(L'pitals more or less)

Answer 44

well i dont think you can apply Lhopital;s easily

Answer 45

whichi is why u use leading coefficients

Answer 46

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

Answer 47

a closed form in terms of n

Answer 48

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

Answer 49

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

Answer 50

wait, can you state the original problem again

Answer 51

sure. just a minute

Answer 52

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

Answer 53

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

Answer 54

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

Answer 55

right that works

Answer 56

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

Answer 57

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