# Computing the double series

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? – k1next Feb 8 '13 at 8:30
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

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