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Put to limits

  1. Dec 31, 2007 #1
    i am not used to so much of maths
    but sometime it makes my life hell when i dont find the answer to my questions
    here one such

    Limit(n[tex]\rightarrow\infty[/tex]){sum(1^p+2^p.......n^p)/n^(p+1)}

    where p is a constant
     
    Last edited: Dec 31, 2007
  2. jcsd
  3. Dec 31, 2007 #2

    CompuChip

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    [tex]\lim_{n \to \infty}\left( \frac{1^p + 2^p + \cdots + n^p}{n^{p+1}}\right) =
    \lim_{n \to \infty} \left( \frac{1^p}{n^{p+1}} + \frac{2^p}{n^{p+1}} + \cdots + \frac{n^p}{n^{p+1}} \right)[/tex]
    What can you say about all the terms separately?
     
  4. Dec 31, 2007 #3
    well i just an hour before solved the question
    but lets see your method

    answer is not another limit, but an simple expression

    but tell me how did you write math expressions on reply,the way you have written
     
    Last edited: Dec 31, 2007
  5. Dec 31, 2007 #4

    CompuChip

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    OK, glad you solved it.
    To get the fancy math just type [tex] and [/ tex] (but without the space between / and tex) and write LaTeX code in between.
     
  6. Dec 31, 2007 #5
    thanks alot...
     
  7. Dec 31, 2007 #6

    Gib Z

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    What was your answer? I know the answer if |p| is greater or equal to 1, and if p=0, but I can't be stuffed to fill in the gaps.
     
  8. Dec 31, 2007 #7
    the answer is true for all real values of p and it is
    1/(P+1).

    try to telescope the whole series .....
     
  9. Dec 31, 2007 #8

    Gib Z

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    That answer is not correct..

    eg for p=-2, the limit becomes [tex]\lim_{n \to \infty}\left( \frac{1^p + 2^p + \cdots + n^p}{n^{p+1}}\right) =
    \lim_{n \to \infty} \left( \frac{1^p}{n^{p+1}} + \frac{2^p}{n^{p+1}} + \cdots + \frac{n^p}{n^{p+1}} \right) = \lim_{n\to \infty} \left( \frac{n}{1^2} + \frac{n}{2^2} + \frac{n}{3^2} ...\right )[/tex] which obviously diverges...
     
  10. Jan 1, 2008 #9
    sorry it was written their that p is the element of R+
    i dont know what that mean

    then i think it is real positive no.
     
    Last edited: Jan 1, 2008
  11. Jan 1, 2008 #10

    Gib Z

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    O that means p is a Positive real number. I still don't see how the series telescopes, perhaps im blind today :(
     
  12. Jan 1, 2008 #11
    try this stuff

    (1+x)^p=???????
    taking x>=1
    rearrange and buy a new telescope to see the sum:rofl:
     
  13. Jan 1, 2008 #12

    Gib Z

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    I'm still lost :(
     
  14. Jan 1, 2008 #13
    ok
    (1+x)^(p+1)=x^(p+1) *(1+1/x)^(p+1)

    (1+x)^(p+1)-x^(p+1)=????????/
    do it taking x having different values from 0 to n
    then add them

    you will get
    (1+x)^(p+1)-1=sum of like terms ...(p+1) * sum(x ^(p+1)) +etc terms

    do you get it
     
    Last edited: Jan 1, 2008
  15. Jan 1, 2008 #14

    Gib Z

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    Ahh ill try this tomorrow, my headache is killing me. I won't be able to say this in an hour, but =D - I haven't slept all YEAR!!
     
  16. Jan 1, 2008 #15
    it is not only a bit complicated but a slight long but i dont get any other method except telescoping(diff. of consequetive terms) to do it without using any formulas or bernoulis method which i think is beyond the standard of a high school guy.
     
  17. Jan 1, 2008 #16

    Gib Z

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    Bernoulli's method? I don't know what you mean, but I did immediately think about using Faulhaber's formula, which gives the sum of the first n k-th powered integers in terms of polynomials whose coefficients are the Bernoulli numbers. I also thought it wouldn't be suitable to apply anything that advanced here either. What exactly is Bernoulli's method?

    EDIT: http://mathworld.wolfram.com/BernoullisMethod.html doesnt seem to relate :(
     
  18. Jan 1, 2008 #17
    what you said is mentioned as bernoulis method in my college library book .
    well thanks for telling me that.
     
  19. Jan 8, 2008 #18

    Gib Z

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    If you are interested, I do have another method =] It seems to be much easier as well =P
     
  20. Jan 9, 2008 #19
    well then please bring it up

    i don't know why this post is not highlighted in the index...........strange
     
  21. Jan 9, 2008 #20

    Gib Z

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    It just so happens that when we wish to evaluate the integral [tex]\int^b_0 x^k dx[/tex], there are two main methods of doing it- Using the Fundamental theorem of Calculus, Or Integration by Riemann sums. The first method of course yields [tex]\frac{b^{k+1}}{k+1}[/tex].

    Now when I proved this to myself the first time, I tried to use equal subdivisions of the integral, which previously had always worked for me. You'll see soon why I ran into a problem using that subdivision =] Instead I had to solve the more general integral, with lower bound a, with the subdivisions [itex]q= \left(\frac{b}{a} \right)^{1/n}[/itex].

    Well anyway, the problem with equal subdivisions meant that when one approximated the integral with n upper Riemann sums, one ran into;

    [tex]\frac{b}{n} \left( (\frac{b}{n})^k + ( \frac{2b}{n})^k + (\frac{3b}{n})^k ....(\frac{nb}{n})^k \right)[/tex]. Taking out the common factor;

    [tex]\frac{b^{k+1}}{n^{k+1}} ( 1^k + 2^k + 3^k + 4^k.....+ n^k)[/tex].

    Now taking limits as n goes to infinity, since we know from the other method that this method should yield the b^(k+1)/(k+1), your limit is k+1. yay
     
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