A community for students.
Here's the question you clicked on:
 0 viewing
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]
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]

This Question is Closed

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0how do you evaluate sum(i^9,i=1..n) do you know?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0please give a little more background. What class? Is this an infinite series?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is there anything else to the question

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0looks like a geometric series

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I have to think in terms of integrals to solve it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you need a power series representation and get the ratio from that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you want (1/n)^10 + (2/n)^10 + ...(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the ratio of consecutive terms is not constant

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can yuo post the original question

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah thats what I asked

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the limit is probably infinity

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its not a decreasing function

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no , if its a riemann sum its area under a curve

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that means it converges

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i just cant see the integral

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well do you see that n is finite there

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0uphill, in that expression

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lim (n>oo) 1/n[(1/n)^9 + (2/n)^9 + (3/n)^9 + ... + (n/n)^9]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0n approaches infinity

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes but each n is finite

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0maybe you need to set up an inequality

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0S sub n + the limit of the integral of that is > the limit (of stuff)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well the denominator is n^9

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no you dont need that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lim (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
 5 years ago
Best ResponseYou've already chosen the best response.0you need to find the sum formula for sum k^9 as for k = 1 to n

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0let me try and make the problem look more neat and type it again

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0when 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
 5 years ago
Best ResponseYou've already chosen the best response.0leading coefficients?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(L'pitals more or less)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well i dont think you can apply Lhopital;s easily

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0whichi is why u use leading coefficients

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we need to find a formula for 1^9 + 2^9 + ... n^9

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0a closed form in terms of n

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay. i know formulas for powers 1,2,3 but not 9

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and thanks you everyone for helping me out on this one. its pretty tough

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait, can you state the original problem again

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{1}x^{9}dx=x^{10}/10, x=0..1=1/10\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{n \rightarrow \infty} (1/n)[(1/n)^9+(2/n)^9+(3/n)^9+...+(n/n)^9]\]

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0deltax=[ba]/n=[10]/n

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0or you can do the limit proof directly , find the sum of k^9

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thank you very much! i finally understand it. well more than i did beforehand anyways.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.