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Limit with Riemann Integral

  1. Jan 20, 2009 #1
    1. The problem statement, all variables and given/known data
    I need to find the following:
    [tex]\lim_{n\rightarrow\infty}\left(\frac{1^2+2^2+3^2+...+n^2}{n^3} \right)[/tex]

    2. Relevant equations



    3. The attempt at a solution
    I know I could do the sum of the series to find the result but I would like to use Riemann sums.
    I think I have to start by writing the limit as:
    [tex]\lim_{n\rightarrow\infty}\sum_{k=1}^{n}\frac{1}{n}\left(\frac{k}{n} \right)^2[/tex]
    However, I'm not sure how to continue.

    Thanks for your help.
     
    Last edited: Jan 20, 2009
  2. jcsd
  3. Jan 20, 2009 #2
    12 + 22 + 32 + ... + n2 = [n(n+1)(2n+1)]/6
     
  4. Jan 20, 2009 #3

    quasar987

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    The thing with Riemann sums is that they are related to the Riemann integral through the equation

    [tex]\lim_{\max\{\Delta x_i\}\rightarrow 0}\sum_{i=1}^n f(x_i^*)\Delta x_i=\int_a^bf(x)dx[/tex]

    This means that for a bounded integrable function f:[a,b]-->R and any sequence [itex]\{\mathcal{P}_n\}_{n\in\mathbb{N}}[/itex] of partitions of [a,b] such that the largest subinterval in [itex]\mathcal{P}_n=\{a=x_0^{(n)}, x_1^{(n)},\ldots,x_N_n^{(n)}=b\}[/itex] goes to zero as [itex]n\rightarrow\infty[/itex], the sequence of associated Riemann sums approaches the integral of f over [a,b]:

    [tex]\left(\lim_{n\rightarrow\infty}\left[\max_{1\leq i\leq N_n}(\Delta x_i^{(n)})\right]= 0 \right)\Longrightarrow \left(\lim_{n\rightarrow\infty}\sum_{i=1}^{N_n}f(x_i^*)\Delta x_i^{(n)} = \int_a^bf(x)dx\right)[/tex]

    So...

    What we have here is the limit

    [tex]\lim_{n\rightarrow\infty}\sum_{i=1}^{n}\frac{1}{n} \left(\frac{i}{n} \right)^2[/tex]

    Notice that for each positive integer n, {0=0/n, 1/n, 2/n ...,(n-1)/n, n/n=1} is a partition of the interval [0,1] and in such a partition, each subinterval has lenght 1/n. And [itex]1/n\rightarrow 0[/itex] as [itex]n\rightarrow\infty[/itex], so according to the big implication above,

    [tex]\lim_{n\rightarrow\infty}\sum_{i=1}^{n}f(x_i^*)\frac{1}{n} = \int_0^1f(x)dx[/tex]

    for any bounded integrable function f:[0,1]-->R.

    Now, for which function f and choice of [itex]x_i^*\in [(i-1)/n,i/n][/itex] do we have

    [tex]f(x_i^*)=\left(\frac{i}{n}\right)^2[/tex]

    ??

    Once you figure that out, you can use the fundamental theorem of calculus to find the value of the integral of f and hence the value of the sum you're interested in.
     
  5. Jan 20, 2009 #4
    OK, so if I understood well x ∈ [0,1] since 1/n is creating partitions of that interval.

    Then we have xi* ∈ [xi-1, xi].
    Since xi - xi-1 = i/n we have xi* = i/n.
    Therefore f(x) = x2.

    Then:
    [tex]\lim_{n\rightarrow\infty}\left(\frac{1^2+2^2+3^2+. ..+n^2}{n^3} \right) = \int_0^1 x^2 dx = \frac{1}{3}[/tex]
    I checked the result and it's correct, what about the proof?

    Thanks again.
     
  6. Jan 20, 2009 #5

    quasar987

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    I'm not sure I agree with your chain of implications from a logical standpoint but you're the one who's best placed to say if you understand or not.

    And what do you mean by "what about the proof"?

    In any case, the way I would have solved the problem is simply by noting that, in view of what has been say above, the sums

    [tex]\sum_{i=1}^{n}\frac{1}{n} \left(\frac{i}{n} \right)^2[/tex]

    are Riemann sums of the function f:[0,1]-->R defined by f(x)=x² where the points [itex]x_i^*\in [x_{i-1},x_i]=[(i-1)/n,i/n][/itex] have been chosen equal to i/n.

    Hence,

    [tex]
    \lim_{n\rightarrow\infty}\sum_{i=1}^{n}\frac{1}{n} \left(\frac{i}{n} \right)^2 =\int_0^1x^2dx=1/3[/tex]
     
  7. Jan 20, 2009 #6
    By the "proof" I meant the fact that the function is f(x) = x2. But now I realize I made a mistake since xi - xi-1 = 1/n ≠ i/n.
    So I don't understand how do you know the function is f(x) = x2?
    Is it because xi* = i/n, and if so, how can you find that?
    Thanks.
     
  8. Jan 20, 2009 #7

    quasar987

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    You do not know a priori that xi* = i/n.

    Suppose more generally that you are given the sum following Riemann sum

    [tex]\sum_{i=1}^{n}\frac{1}{n}F(i,n) \right)[/tex]

    where F(i,n) is some expression involving i and n.

    The question is,

    "For which function f:[0,1]-->R and which choice of points [itex]x_i^*\in [(i-1)/n,i/n][/itex] do we have [itex]f(x_i^*)=F(i,n)[/itex] for all i=0,...,n ??"

    In the present case, F(i,n)=(i/n)² and so after a moment of reflection trying various combinations of functions and points [itex]x_i^*[/itex], you will conclude that f(x)=x² and [itex]x_i^*=i/n[/itex] does the trick.
     
  9. Jan 20, 2009 #8
    Thanks, I think I understood.

    So would this be correct so far?
    [tex]\lim_{n\rightarrow\infty}\left(\frac{1}{\sqrt{n^2+n}} + \frac{1}{\sqrt{n^2+2n}}+\frac{1}{\sqrt{n^2+3n}}+...+\frac{1}{\sqrt{n^2+nn}} \right) = \lim_{n\rightarrow\infty}\sum_{i=1}^{n}\frac{1}{\sqrt{n}}\frac{1}{\sqrt{n+i}} = \lim_{n\rightarrow\infty}\sum_{i=1}^{n}\frac{1}{\sqrt{n}}f(x_i^*)=\int_{0}^{\infty}f(x)dx[/tex]

    With [itex]x_i^*\in[(i-1)/\sqrt{n}, i/\sqrt{n}][/itex].
    I thought the interval is [0, ∞) since for each n, we have {0, 1/√n, 2/√n, ..., n/√n} a partition of [0, ∞), is that right?
    Now I would need to find xi* and f just by guessing?
     
  10. Jan 20, 2009 #9

    quasar987

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    The theorem concerns linking Riemann sum and integral concerns definite integrals. [0,∞) is not a valid interval.

    But

    [tex]\frac{1}{\sqrt{n^2+in}}=\frac{1}{n}\frac{1}{\sqrt{1+i/n}}[/tex]
     
  11. Jan 21, 2009 #10
    OK, now it all makes sense :)
    [tex]\lim_{n\rightarrow\infty}\left(\frac{1}{\sqrt{n^2+ n}} + \frac{1}{\sqrt{n^2+2n}}+\frac{1}{\sqrt{n^2+3n}}+...+\frac{1}{\sqrt{n^2+nn}} \right) = \lim_{n\rightarrow\infty}\sum_{i=1}^{n}\frac{1}{n}\frac{1}{\sqrt{1+\frac{i}{n}}} =\int_{0}^{1}\frac{dx}{\sqrt{1+x}}=\left[ 2\sqrt{1+x}\right]_0^1=2\sqrt{2}-2[/tex]
     
  12. Jan 21, 2009 #11

    quasar987

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    Well done.
     
  13. Jan 21, 2009 #12
    I don't mean to hijack this thread, but I have a question for quasar.

    In this step right here: [tex] \lim_{n\rightarrow\infty}\sum_{i=1}^{n}\frac{1}{n} \frac{1}{\sqrt{1+\frac{i}{n}}} =\int_{0}^{1}\frac{dx}{\sqrt{1+x}}=\left[ 2\sqrt{1+x}\right]_0^1[/tex],

    Is the integral from 0 to 1 because in the summation, when i = 1 then the term 1/n = 1 and as i tends to infinity, the term 1/n will tend to 0?
     
  14. Jan 21, 2009 #13

    quasar987

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    The integral is from 0 to 1 because i/n is some point taken from the subinterval [(i-1)/n, i/n] of the partition {0, 1/n, 2/n, ..., n/n=1} of [0,1]. It's all explained in post #3.
     
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