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I need a starting point for

$$\lim_{n\to\infty}\sum_{i=1}^{n}\sum_{j=1}^{n}\frac{\ln\left(1+\displaystyle \frac{i}{n}\right)\ln\left(1+\displaystyle\frac{j}{n}\right)}{\sqrt{n^4+i^2+j^2}}$$ What would you suggest me to do? Thanks!

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Do you want to know something about convergence, an analytical solution or a numerical solution? –  sonystarmap Feb 8 '13 at 8:30
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Is it really $i^2$ and $j^2$ in $\sqrt{n^4 + i^2 + j^2}$? If yes, they drop out from the final limit. The limit becomes a square of a Riemann sum... –  achille hui Feb 8 '13 at 8:38

1 Answer 1

Well, if the denominator of the summand is correct, it will approach $1/n^2$ as $n \rightarrow \infty$. The result is just a pair of Riemann sums corresponding to

$$\left ( \int_0^1 dx \: \log(1+x) \right )^2 = (2 \log{2}-1)^2$$

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Of course, one really should do things properly; e.g. rewrite the summand as $$ \frac{1}{n^2} \ln\left(1 + \frac{i}{n}\right) \ln\left(1 + \frac{j}{n}\right) + \ln\left(1 + \frac{i}{n}\right) \ln\left(1 + \frac{j}{n}\right) \left( \frac{1}{\sqrt{n^4 + i^2 + j^2}} - \frac{1}{n^2} \right) $$ and show the sum of the second term really does converge to 0. I imagine the second term works out to $O(n^{-4})$ which is good enough. –  Hurkyl Feb 8 '13 at 9:03
    
@Hurkyl: How is what I did improper? $i$ and $j$ are each at most $n$, so by factoring out the $n^4$ from the denominator, we get $1/n^2$ in the desired limit. I do not know how your rearrangement of the terms provides any additional insight or rigor. –  Ron Gordon Feb 8 '13 at 9:08
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@Hurkyl: sometimes, sure. Not here. Whatever errors were introduced by my approximation were also introduced by yours...because they are the same approximation in the limit. Viz., $(n^4+i^2+j^2)^{-1/2} \sim 1/n^2 [1 - 1/(2 n^2) ((i/n)^2 + (j/n)^2) + O(1/n^4)]$ Where is the error in the limit that your rearrangement fixes? –  Ron Gordon Feb 8 '13 at 9:21
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There is no error in my rearrangement; it's exactly equal to the summand. There are a number of ways to bound the error; rearrangements like I did above are a common starting point, and so replacing the term being approximated with a big-O simplification. I'm happy to assume you are also experienced enough to have a reliable intuition that the error will vanish without having to bother with the calculation, but for those less experienced do not have reliable intuition, and may not even be aware of the potential problem. For them, a demonstration (or at least acknowledgement) should be given. –  Hurkyl Feb 8 '13 at 9:38
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@Hurkyl: I'll grant you that a more in-depth computation of the error is warranted when the OP needs some hand-holding. I do appreciate the exact splitting into the main term and the error, although, again, our representations produce the same error. Chris'sSister, on the other hand, shares our intuition and in that case, I felt it sufficient to bring her around to the main point without the added explanation. –  Ron Gordon Feb 8 '13 at 9:44

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