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Let $\displaystyle u_n =\frac{n}{2}-\sum_{k=1}^n\frac{n^2}{(n+k)^2}$. The question is: Find the limit of the sequence $(u_n)$. The problem is if we write $\displaystyle u_n=n\left(\frac{1}{2}-\frac{1}{n}\sum_{k=1}^n\frac{1}{(1+\frac{k}{n})^2}\right)$ and we use the fact that the limit of Riemann sum $\displaystyle \frac{1}{n}\sum_{k=1}^n\frac{1}{(1+\frac{k}{n})^2}$ is $\displaystyle \int_0^1 \frac{dx}{(1+x)^2}=\frac{1}{2}$ we find the indeterminate form $\infty\times 0$. How can we avoid this problem? Thanks for help.

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3 Answers 3

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Here's a rigorous derivation.

$$ \frac{n}{2} - \sum_{k=1}^n \frac{n^2}{(n+k)^2} = n \int_0^1 \frac{dx}{(1+x)^2} - \sum_{k=1}^n \frac{1}{(1+\frac{k}{n})^2} \\ = \sum_{k=1}^n \left(\int_{\frac{k-1}{n}}^{\frac{k}{n}} \frac{ n\, dx}{(1+x)^2} - \frac{1}{(1+\frac{k}{n})^2}\right) \\ = \sum_{k=1}^n \left(\int_{-\frac{1}{n}}^0 \frac{ n\, dx}{(1+\frac{k}{n} + x)^2} - \frac{1}{(1+\frac{k}{n})^2}\right) \\ = \sum_{k=1}^n \left(n\int_{-\frac{1}{n}}^0 \left(\frac{1}{(1+\frac{k}{n})^2} - \frac{2x}{(1+\frac{k}{n})^3} + \frac{3 x^2_*(x,n,k)}{(1+\frac{k}{n} +x_*(x,n,k))^4}\right)dx - \frac{1}{(1+\frac{k}{n})^2}\right)\\ = \sum_{k=1}^n n \int_{-\frac{1}{n}}^0 \left(- \frac{2x}{(1+\frac{k}{n})^3} + \frac{3 x^2_*(x,n,k)}{(1+\frac{k}{n}+x_*(x,n,k))^4}\right)dx. $$ In the last two lines, we've used Taylor's theorem for $(1+\frac{k}{n}+x)^{-2}$. By that theorem with the Lagrange form of the remainder, $x_*(x,n,k)$ is some point in $(-\frac{1}{n},0)$. Given this bound on $x_*$, it is easy to see the term involving $x_*$ is $O(n^{-1})$, so we can ignore it in the limit $n\to\infty$. Continuing, we evaluate the limit of the remaining term: $$ \lim_{n\to\infty} \sum_{k=1}^n n \int_{-\frac{1}{n}}^0 \left(- \frac{2x}{(1+\frac{k}{n})^3} \right)dx = \lim_{n\to\infty} \sum_{k=1}^n n \left(\frac{1}{n^2(1+\frac{k}{n})^3} \right)dx \\ = \int_0^1 \frac{dx}{(1+x)^3} = \frac{3}{8}. $$

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    $\begingroup$ Is your integral correct? $\endgroup$ Feb 21, 2013 at 11:20
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    $\begingroup$ @Ishan - The last integral is correct.Which integral are you questioning? $\endgroup$ Feb 21, 2013 at 16:54
  • $\begingroup$ What $x_*(x,n,k)$ means ? $\endgroup$
    – ParaH2
    Jul 10, 2015 at 8:37
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Not sure if this is correct, but here goes anyway,

$$u_n=n\left(\int_0^1\frac{1}{(1+x)^2}dx-\frac 1n\sum_{k=1}^n\frac{1}{(1+\frac{k}{n})^2}\right)$$

Let $$\begin{align} A_k &=\int_{\frac{k-1}{n}}^{\frac{k}{n}}\frac{1}{(1+x)^2}\mathrm{d}x -\frac{1}{n\left(1+\frac{k}{n}\right)^2} \\ & =\frac{1}{n\left(1+\frac{k}{n}\right)}\cdot\left(\frac{1}{1+\frac{k-1}{n}}-\frac{1}{1+\frac{k}{n}}\right) \\ &=\frac{1}{n^2}\cdot\frac{1}{\left(1+\frac kn\right)^2}\frac{1}{1+\frac{k-1}{n}} \end{align}$$

Then $$u_n=\sum_{k=1}^nnA_k=\int_0^1\frac{1}{(1+x)^3}dx=\frac 38$$

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    $\begingroup$ Nice! Looks correct to me. $\endgroup$
    – Alex Zorn
    Feb 21, 2013 at 16:55
  • $\begingroup$ Very nice solution - better than mine. To make it rigorous, you just need some justification for passing from the sum to the integral in the last line, but that's trivial (dominated convergence, for example). $\endgroup$ Feb 21, 2013 at 19:04
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$$\dfrac1{(1+k/n)^2} = \left( 1 + \dfrac{k}n \right)^{-2} = 1 + \dfrac{(-2)}{1!} \dfrac{k}n + \dfrac{(-2)(-3)}{2!} \dfrac{k^2}{n^2} + \dfrac{(-2)(-3)(-4)}{3!} \dfrac{k^3}{n^3} + \cdots $$ Now recall that $$\sum_{k=1}^n \left(\dfrac{k}n \right)^m = \dfrac{n}{m+1} + \dfrac12 + \mathcal{O}(1/n)$$ Hence, \begin{align} \sum_{k=1}^n \dfrac1{(1+k/n)^2} & = n + \dfrac{(-2)}{1!} \left(\dfrac{n}2 + \dfrac12 + \mathcal{O}(1/n)\right)\\ & + \dfrac{(-2)(-3)}{2!} \left(\dfrac{n}3 + \dfrac12 + \mathcal{O}(1/n)\right)\\ & + \dfrac{(-2)(-3)(-4)}{3!} \left(\dfrac{n}4 + \dfrac12 + \mathcal{O}(1/n)\right)\\ & + \dfrac{(-2)(-3)(-4)(-5)}{4!} \left(\dfrac{n}5 + \dfrac12 + \mathcal{O}(1/n)\right)\\ & + \cdots \end{align} The leading order term is $n-n+n-n+n-n + \cdots$, which when regularized gives us $\dfrac{n}2$. The next term is $$\dfrac12 \left(-2+3-4+5-6 \pm \cdots\right)$$ which when regularized gives us $$\dfrac{-1+\eta(-1)}2 = \dfrac{-1+\dfrac14}2 = - \dfrac38$$ Hence, $$\sum_{k=1}^n \left(1+\dfrac{k}n\right)^{-2} = \dfrac{n}2 - \dfrac38 + \mathcal{O}(1/n)$$ Hence, the limit is $\dfrac38$.

The argument, though it appears to be non-rigorous, can be made rigorous with careful regularization.

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  • $\begingroup$ thanks for answer, but what do you mean by regularize the sum $\displaystyle\sum_{k=1}^{\infty}(-1)^k $? Do you mean we take the average of this sum? $\endgroup$
    – user63181
    Feb 21, 2013 at 9:12
  • $\begingroup$ @sbr You may want to look here for details on regularizing sums. $\endgroup$
    – user17762
    Feb 21, 2013 at 9:23
  • $\begingroup$ Ok thanks, the regularized sum term is known in the french jargon as Cezaro average. $\endgroup$
    – user63181
    Feb 21, 2013 at 10:25
  • $\begingroup$ @sbr Cezaro way is one way of regularization. There are many other ways to regularize, though interestingly all these give the same answer for the regularized sum. $\endgroup$
    – user17762
    Feb 21, 2013 at 19:50

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