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Integrating for approximation of a sum

  1. May 15, 2014 #1
    1. The problem statement, all variables and given/known data

    Find an N so that ##∑^{\infty}_{n=1}\frac{log(n)}{n^2}## is between ##∑^{N}_{n=1}\frac{log (n)}{n^2}## and ##∑^{N}_{n=1}\frac{log(n)}{n^2}+0.005.##

    2. Relevant equations
    Definite integration


    3. The attempt at a solution
    I began by taking a definite integral: ##\int^{\infty}_{N}\frac{log(n)}{n^2}dn## and, using integration by parts, arrived at the following answer: ##\frac{log(N)+1}{N}##. (Is this right? If not, I could post the steps I used to try to see where I made an error)

    Next we need ##\frac{log(N)+1}{N}## to be within 0.005 as given by the problem, so:

    ##\frac{log(N)+1}{N}=0.005=\frac{1}{200}##

    But I'm having trouble how to solve for N algebraically. Would appreciate any help.
     
  2. jcsd
  3. May 15, 2014 #2

    BiGyElLoWhAt

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    Gold Member

    I actually got -1 times what you have. remember: ##\int \frac{1}{n^2} = -\frac{1}{n}##
     
  4. May 15, 2014 #3
    Right but since we're taking a definite integral on the interval from N to ∞, and since ##\frac{-log(N)-1}{N}## as N approaches infinity equals zero, shouldn't the definite integral work out to:
    ##0-\frac{-log(N)-1}{N}=\frac{log(N)+1}{N}##?
     
  5. May 15, 2014 #4

    pasmith

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    Homework Helper

    Your integral is correct.

    The graph of [itex]x^{-2}\log(x)[/itex] has a maximum in [itex][1,2][/itex], so it's difficult to say whether the integral is an over-estimate or under-estimate of the sum. To be safe I'd require that [tex]\frac{\log(N) + 1}{N} < \frac{1}{400}[/tex].

    Equations of that type have to be solved numerically. You're looking for zeroes in [itex]x > 0[/itex] of [tex]f(x) = \frac{x}{200} - 1 - \log x[/tex]. The derivative of this function is [tex]
    f'(x) = \frac{1}{200} - \frac{1}{x}[/tex] so the second derivative ([itex]x^{-2}[/itex]) is everywhere positive. There is a minimum at [itex]x = 200[/itex] where [itex]f(200) < 0[/itex]. Since [itex]f(x) \to +\infty[/itex] as [itex]x \to 0[/itex] or [itex]x \to \infty[/itex] there are exactly two solutions. Logic dictates you want the larger, since if [itex]N[/itex] works then any larger [itex]M[/itex] should also work.
     
  6. May 15, 2014 #5
    Thanks, I tried solving it numerically and came up with N = 1687, which the textbook confirms is correct.
     
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