# Limit of a sequence with indeterminate form

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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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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Is your integral correct? –  Ishan Banerjee Feb 21 '13 at 11:20
@Ishan - The last integral is correct.Which integral are you questioning? –  Will Nelson Feb 21 '13 at 16:54

Not sure if this is correct, but here goes anyway,

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

Let $A_k=\int_{\frac{k-1}{n}}^{\frac{k}{n}}\frac{1}{(1+x)^2}dx -\frac{1}{n(1+\frac{k}{n})^2}=\frac{1}{n(1+\frac{k}{n})}(\frac{1}{1+\frac{k-1}{n}}-\frac{1}{1+\frac{k}{n}})=\frac{1}{n^2}.\frac{1}{(1+\frac kn)^2}.\frac{1}{1+\frac{k-1}{n}}$

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

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Nice! Looks correct to me. –  Alex Zorn Feb 21 '13 at 16:55
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). –  Will Nelson Feb 21 '13 at 19:04

$$\dfrac1{(1+k/n)^2} = \left( 1 + \dfrac{k}n \right)^{-2} = 1 + \dfrac{(-2)}{1!} \dfrac{k}n + \dfrac{(-2)(-3)}{2!} \dfrac{k^2}{n^2} + \dfrac{(-2)(-3)(-4)}{3!} \dfrac{k^3}{n^3} + \cdots$$ 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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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? –  Sami Ben Romdhane Feb 21 '13 at 9:12
@sbr You may want to look here for details on regularizing sums. –  user17762 Feb 21 '13 at 9:23
Ok thanks, the regularized sum term is known in the french jargon as Cezaro average. –  Sami Ben Romdhane Feb 21 '13 at 10:25
@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. –  user17762 Feb 21 '13 at 19:50