## anonymous 5 years ago what is the limit as n approaches infinity of 1/n[(1/n)^9 + (2/n)^9 + (3/n)^9 + ... + (n/n)^9]

1. myininaya

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

2. anonymous

no

3. anonymous

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

4. anonymous

is there anything else to the question

5. anonymous

looks like a geometric series

6. anonymous

^yeah

7. anonymous

so find the ratio

8. anonymous

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

9. anonymous

why?

10. anonymous

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

11. anonymous

the ratio is 2

12. anonymous

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

13. anonymous

i take that back

14. anonymous

the ratio of consecutive terms is not constant

15. anonymous

right

16. anonymous

this is an integral

17. anonymous

can yuo post the original question

18. anonymous

19. anonymous

oh its a riemann sum

20. anonymous

the limit is probably infinity

21. anonymous

its not a decreasing function

22. anonymous

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

23. anonymous

that means it converges

24. anonymous

i just cant see the integral

25. anonymous

well do you see that n is finite there

26. anonymous

uphill, in that expression

27. anonymous

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

28. anonymous

n approaches infinity

29. anonymous

yes but each n is finite

30. anonymous

this is not a series

31. anonymous

this is a rieman sum

32. anonymous

maybe you need to set up an inequality

33. anonymous

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

34. anonymous

well the denominator is n^9

35. anonymous

no you dont need that

36. 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 ]

37. anonymous

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

38. anonymous

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

39. anonymous

give me medal

40. anonymous

i solve it

41. 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?

42. anonymous

43. anonymous

(L'pitals more or less)

44. anonymous

well i dont think you can apply Lhopital;s easily

45. anonymous

whichi is why u use leading coefficients

46. anonymous

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

47. anonymous

a closed form in terms of n

48. anonymous

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

49. anonymous

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

50. anonymous

wait, can you state the original problem again

51. anonymous

sure. just a minute

52. myininaya

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

53. anonymous

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

54. myininaya

55. myininaya

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

56. anonymous

right that works

57. anonymous

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

58. anonymous

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