anonymous
  • anonymous
what is the limit as n approaches infinity of 1/n[(1/n)^9 + (2/n)^9 + (3/n)^9 + ... + (n/n)^9]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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myininaya
  • myininaya
how do you evaluate sum(i^9,i=1..n) do you know?
anonymous
  • anonymous
no
anonymous
  • anonymous
please give a little more background. What class? Is this an infinite series?

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anonymous
  • anonymous
is there anything else to the question
anonymous
  • anonymous
looks like a geometric series
anonymous
  • anonymous
^yeah
anonymous
  • anonymous
so find the ratio
anonymous
  • anonymous
I have to think in terms of integrals to solve it.
anonymous
  • anonymous
why?
anonymous
  • anonymous
you need a power series representation and get the ratio from that
anonymous
  • anonymous
the ratio is 2
anonymous
  • anonymous
you want (1/n)^10 + (2/n)^10 + ...(
anonymous
  • anonymous
i take that back
anonymous
  • anonymous
the ratio of consecutive terms is not constant
anonymous
  • anonymous
right
anonymous
  • anonymous
this is an integral
anonymous
  • anonymous
can yuo post the original question
anonymous
  • anonymous
yeah thats what I asked
anonymous
  • anonymous
oh its a riemann sum
anonymous
  • anonymous
the limit is probably infinity
anonymous
  • anonymous
its not a decreasing function
anonymous
  • anonymous
no , if its a riemann sum its area under a curve
anonymous
  • anonymous
that means it converges
anonymous
  • anonymous
i just cant see the integral
anonymous
  • anonymous
well do you see that n is finite there
anonymous
  • anonymous
uphill, in that expression
anonymous
  • anonymous
lim (n->oo) 1/n[(1/n)^9 + (2/n)^9 + (3/n)^9 + ... + (n/n)^9]
anonymous
  • anonymous
n approaches infinity
anonymous
  • anonymous
yes but each n is finite
anonymous
  • anonymous
this is not a series
anonymous
  • anonymous
this is a rieman sum
anonymous
  • anonymous
maybe you need to set up an inequality
anonymous
  • anonymous
S sub n + the limit of the integral of that is > the limit (of stuff)
anonymous
  • anonymous
well the denominator is n^9
anonymous
  • anonymous
no you dont need that
anonymous
  • 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
  • anonymous
you need to find the sum formula for sum k^9 as for k = 1 to n
anonymous
  • anonymous
let me try and make the problem look more neat and type it again
anonymous
  • anonymous
give me medal
anonymous
  • anonymous
i solve it
anonymous
  • 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
  • anonymous
leading coefficients?
anonymous
  • anonymous
(L'pitals more or less)
anonymous
  • anonymous
well i dont think you can apply Lhopital;s easily
anonymous
  • anonymous
whichi is why u use leading coefficients
anonymous
  • anonymous
we need to find a formula for 1^9 + 2^9 + ... n^9
anonymous
  • anonymous
a closed form in terms of n
anonymous
  • anonymous
okay. i know formulas for powers 1,2,3 but not 9
anonymous
  • anonymous
and thanks you everyone for helping me out on this one. its pretty tough
anonymous
  • anonymous
wait, can you state the original problem again
anonymous
  • anonymous
sure. just a minute
myininaya
  • myininaya
\[\int\limits_{0}^{1}x^{9}dx=x^{10}/10, x=0..1=1/10\]
anonymous
  • anonymous
\[\lim_{n \rightarrow \infty} (1/n)[(1/n)^9+(2/n)^9+(3/n)^9+...+(n/n)^9]\]
myininaya
  • myininaya
1 Attachment
myininaya
  • myininaya
deltax=[b-a]/n=[1-0]/n
anonymous
  • anonymous
right that works
anonymous
  • anonymous
or you can do the limit proof directly , find the sum of k^9
anonymous
  • anonymous
thank you very much! i finally understand it. well more than i did beforehand anyways.

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