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Determination of limit in Riemann sums.

  1. May 19, 2014 #1
    How could I find the lim as n-> infinity of the expression I attached?


    The only way I could find was to express it in terms of a definite integral.

    Integral of xe^(-2x) from 0 to 1.

    What is the other way?
     

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  2. jcsd
  3. May 19, 2014 #2

    jbunniii

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    Have you tried summation by parts? Try setting ##a_i = i##, ##b_i = e^{-2i/n}##, and ##B_i = \sum_{j=1}^i b_j##. Then your sum becomes
    $$\begin{align}
    \frac{1}{n^2}\sum_{i=1}^n a_i b_i
    &= \frac{1}{n^2} \left(a_n b_n - \sum_{i=1}^{n-1}B_i (a_{i+1}-a_i)\right) \end{align}$$
    which should be easier to work with since ##a_{i+1} - a_i = 1## and ##B_i## is a partial sum of a geometric series.
     
  4. May 19, 2014 #3

    HallsofIvy

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    Since you posted another problem with "Riemann sums", yes, that is probably what you are expected to do. I doubt there is any "other way" to do this.

    Now, do you know how, without using "Riemann sums", to find [itex]\int_0^1 xe^{-2x} dx[/tex]?

    (Hint- let u= -2x.)
     
  5. May 19, 2014 #4
    Jbunnii, i followed the site. Apparently that method is similar to integration by parts and it looks quite complicated!
    I think this clarified that i must aimply use the integral which is quite easy to do with integration by parts!
    Thanks for the help!
     
  6. May 19, 2014 #5

    jbunniii

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    [edited to fix a few typos]

    I agree, the integral/Riemann sum method should be the easiest way to solve this, and most likely it's what the author of the question intended.

    The summation by parts approach also works, though. It has the advantage of being more elementary (you don't need to bring in the machinery of Riemann integration) but it does require a bit more work. Here's a sketch. As I wrote above, we set ##a_i = i##, ##b_i = e^{-2i/n}##, and ##B_i = \sum_{j=1}^{i} b_j##. Then we have to evaluate the limit of this sum:
    $$\begin{align}
    \frac{1}{n^2}\sum_{i=1}^n a_i b_i
    &= \frac{1}{n^2} \left(a_n b_n - \sum_{i=1}^{n-1}B_i (a_{i+1}-a_i)\right) \end{align}$$
    Note that ##a_{i+1} - a_i = 1## and
    $$B_i = \sum_{j=1}^{i} e^{-2j/n} = \sum_{j=1}^{i} (e^{-2/n})^j = e^{-2/n}\sum_{j=0}^{i-1}(e^{-2/n})^j = e^{-2/n}\frac{1 - e^{-2i/n}}{1 - e^{-2/n}}$$
    So our sum reduces to
    $$\frac{1}{n^2} \left(n e^{-2} - \frac{e^{-2/n}}{1 - e^{-2/n}} \sum_{i=1}^{n-1} (1 - e^{-2i/n})\right)$$
    Now you need to simplify further and take the limit as ##n \rightarrow \infty##.
     
  7. May 20, 2014 #6
    This too can be handled without the definite integrals. First notice that:
    $$\sum_{i=1}^{n} x^i=x\frac{x^n-1}{x-1}$$
    (You should be able to prove the above as it is a simple geometric progression).

    Differentiate both the sides with respect to ##x## to obtain:
    $$\sum_{i=1}^n i\,x^{i-1}=\frac{nx^{n+1}-(n+1)x^n+1}{(x-1)^2}$$
    $$\Rightarrow \sum_{i=1}^n i\,x^{i}=\frac{nx^{n+2}-(n+1)x^{n+1}+x}{(x-1)^2}$$
    Replace ##x## with ##e^{-2/n}## and divide by ##n^2## on both the sides to obtain:
    $$\sum_{i=1}^n \frac{ie^{-2i/n}}{n^2}=\frac{ne^{-2-4/n}-(n+1)e^{-2-2/n}+e^{-2/n}}{(e^{-2/n}-1)^2n^2}=\frac{n+e^{2/n+2}-e^{2/n}(n+1)}{e^2(e^{2/n}-1)^2n^2}$$
    $$\Rightarrow \sum_{i=1}^n \frac{ie^{-2i/n}}{n^2}=\frac{-n(e^{2/n}-1)+e^{2/n}(e^2-1)}{e^2(e^{2/n}-1)^2n^2}=\frac{-2\left(\frac{e^{2/n}-1}{2/n}\right)+e^{2/n}(e^2-1)}{4e^2\left(\frac{e^{2/n}-1}{2/n}\right)^2}$$
    Notice that as ##n\rightarrow \infty##, ##e^{2/n} \rightarrow 1## and ##\frac{e^{2/n}-1}{2/n} \rightarrow 1##, hence,
    $$\lim_{n\rightarrow \infty}\frac{-2\left(\frac{e^{2/n}-1}{2/n}\right)+e^{2/n}(e^2-1)}{4e^2\left(\frac{e^{2/n}-1}{2/n}\right)^2}=\frac{-2+e^2-1}{4e^2}=\boxed{\dfrac{e^2-3}{4e^2}}$$

    Moral: It is better to use definite integrals. :tongue:
     
  8. May 20, 2014 #7

    D H

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    There's a key error here: That ##a_nb_n## should be ##a_nB_n##. This should be
    [tex]\frac{1}{n^2}\sum_{i=1}^n a_i b_i
    = \frac{1}{n^2} \left(a_n B_n - \sum_{i=1}^{n-1}B_i (a_{i+1}-a_i)\right) [/tex]
     
  9. May 20, 2014 #8

    jbunniii

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    Good catch, thanks!
     
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