# Find $\lim_{n\to \infty}\frac{1}{\ln n}\sum_{j,k=1}^{n}\frac{j+k}{j^3+k^3}.$

Find

$$\lim_{n\to \infty}\frac{1}{\ln n}\sum_{j,k=1}^{n}\frac{j+k}{j^3+k^3}\;.$$

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So no activity on this question for a day, but there are valid answers (in particular, mine :). Are the answers not clear? If not, please let me know where you are not clear and I can improve my answer at least. – Will Nelson Feb 19 '13 at 7:24

$$\frac{1}{\log n}\sum_{j,k=1}^n \frac{j+k}{j^3 +k^3} = \frac{1}{\log n}\sum_{m=1}^n\left(2\sum_{i=1}^m \frac{m+i}{m^3+i^3} - \frac{1}{m^2}\right)$$ Since $$\lim_{n\to\infty} \frac{1}{\log n}\sum_{m=1}^n \frac{1}{m^2} = 0,$$ we conclude $$\lim_{n\to\infty} \frac{1}{\log n}\sum_{j,k=1}^n \frac{j+k}{j^3 +k^3} = 2 \lim_{n\to\infty}\frac{1}{\log n}\sum_{m=1}^n\left(\sum_{i=1}^m \frac{m+i}{m^3+i^3}\right).$$ Define $$a_m = m \cdot \sum_{i=1}^m \frac{m+i}{m^3+i^3},$$ so $$(1)\lim_{n\to\infty} \frac{1}{\log n}\sum_{j,k=1}^n \frac{j+k}{j^3 +k^3} = \lim_{n\to\infty} 2 \frac{1}{\log n} \sum_{m=1}^n \frac{a_m}{m}.$$ We can approximate $a_m$ using a Riemann integral, giving $$(2) \lim_{m\to\infty} a_m = \int_0^1 \frac{1+x}{1+x^3} = \frac{2\pi}{3\sqrt{3}}.$$ (I forget how to do that last integral. I used Mathematica.) Recall that $$\lim_{n\to\infty} \frac{1}{\log n}\sum_{m=1}^n \frac{1}{m} = 1.$$ For any sequence $b_m\to 0$, $$(3) \lim_{n\to\infty} \frac{1}{\log n}\sum_{m=1}^n b_m \frac{1}{m} = 0.$$ To see this, suppose $|b_m|<\epsilon$ for $m>N$. Then $$\left|\frac{1}{\log n}\sum_{m=1}^n b_m \frac{1}{m}\right| <= \left|\frac{1}{\log n}\sum_{m=1}^N b_m \frac{1}{m}\right| + \left|\frac{1}{\log n}\sum_{m>N}^n b_m \frac{1}{m}\right| \\ <= \left|\frac{1}{\log n}\sum_{m=1}^N b_m \frac{1}{m}\right| + \frac{\epsilon}{\log n}\sum_{m>N}^n \frac{1}{m}.$$ The limit of the right hand side of this inequality is $\epsilon$. Since $\epsilon$ can be chosen arbitrarily small, (3) follows. As an immediate corollary, if $b_m\to L$, then $$(4) \lim_{n\to\infty} \frac{1}{\log n}\sum_{m=1}^n b_m \frac{1}{m} = L.$$ It follows from (1), (2), and (4) that $$\lim_{n\to\infty} \frac{1}{\log n}\sum_{j,k=1}^n \frac{j+k}{j^3 +k^3} = \frac{4\pi}{3\sqrt{3}}.$$

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$\sum_{i=1}^{\infty}\frac{1}{i^2}=\frac{{\pi}^2}{6}$ – Ishan Banerjee Feb 18 '13 at 7:19
How can you use the limit of a(m) instead of a(m)? – Ishan Banerjee Feb 18 '13 at 7:27
As far as using the limit of a(m), I think I'll have to edit for clarity. I'm unable to say why in just a sentence or two. – Will Nelson Feb 18 '13 at 7:35

I will be very non-rigorous here.

Letting $m = j+k$,

\begin{align} \sum_{j,k=1}^{n}\frac{j+k}{j^3+k^3} &\approx \frac{1}{2}\sum_{m=2}^{2n} \sum_{j=1}^{m-1}\frac{m}{j^3+(m-j)^3}\\ &= \frac{1}{2}\sum_{m=2}^{2n} m\sum_{j=1}^{m-1}\frac{1}{(j+(m-j))(j^2-j(m-j)+(m-j)^2)}\\ &= \frac{1}{2}\sum_{m=2}^{2n} m\sum_{j=1}^{m-1}\frac{1}{m(j^2-jm+j^2+m^2-2jm+j^2)}\\ &= \frac{1}{2}\sum_{m=2}^{2n} \sum_{j=1}^{m-1}\frac{1}{3j^2-3jm+m^2}\\ &= \frac{1}{2}\sum_{m=2}^{2n} \frac{1}{m^2}\sum_{j=1}^{m-1}\frac{1}{3(j/m)^2-3(j/m)+1}\\ &\approx \frac{1}{2}\sum_{m=2}^{2n} \frac{1}{m}\int_{0}^{1} \frac{dx}{3x^2-3x+1}\\ &= \frac{1}{2}\int_{0}^{1}\frac{dx}{3x^2-3x+1} \sum_{m=2}^{2n} \frac{1}{m}\\ &\approx \frac{\ln(2n)}{2} \int_{0}^{1}\frac{dx}{x^2-x+1/3} \\ \end{align}.

To evaluate the integral,

\begin{align} I = \int_{0}^{1}\frac{dx}{x^2-x+1/3} &=\int_{0}^{1}\frac{dx}{x^2-x+1/4-1/4+1/3}\\ &=\int_{0}^{1}\frac{dx}{(x-1/2)^2+1/12}\\ &=\int_{-1/2}^{1/2}\frac{dx}{x^2+1/12}\\ &=2\int_{0}^{1/2}\frac{dx}{x^2+1/12}\\ \end{align}.

Letting $x = y/\sqrt{12}= y/(2\sqrt{3})$ \begin{align} I &= 2\int_{0}^{1/2}\frac{dx}{x^2+1/12}\\ &= 2\int_{0}^{\sqrt{3}}\frac{dy}{2\sqrt{3}((y^2)/12+1/12)}\\ &= \frac{12}{\sqrt{3}}\int_{0}^{\sqrt{3}}\frac{dy}{y^2+1}\\ &= \frac{12}{\sqrt{3}}\tan^{-1}(\sqrt{3})\\ &= \frac{12}{\sqrt{3}}(\pi/3)\\ &=\frac{4\pi}{\sqrt{3}}\\ &=\frac{4\sqrt{3}\pi}{3}\\ \end{align}

so the sum is about $$\frac{\ln(2n)}{2} \frac{4\sqrt{3}\pi}{3} =\frac{4\sqrt{3}\pi\ln(2n)}{6} \approx \frac{2\sqrt{3}\pi\ln(n)}{3}$$

so the limit is $\dfrac{2\sqrt{3}\pi}{3}$.

As usual, this is done off the top of my head, doing math as $\LaTeX$, but at least I got a reasonable limit. Who knows, this might even be correct.

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why is there a factor of 1/2 from beginning? – Maesumi Feb 18 '13 at 14:29
Oops. I think that was a mistake. I'll fix it later. Thanks. – marty cohen Feb 18 '13 at 16:18

Here is another sketch of proof.

Let

$$J_n = \{(j, k) : 0 \leq j, k < n \text{ and } (j, k) \neq (0, 0) \}.$$

Then for each $(j, k) \in J_n$ and $(x_0, y_0) = (j/n, k/n)$, we have

$$\frac{x_0 + y_0}{(x_0+\frac{1}{n})^{3} + (y_0 + \frac{1}{n})^3} \leq \frac{x+y}{x^3 + y^3} \leq \frac{x_0 + y_0 + \frac{2}{n}}{x_0^3 + y_0^3}$$

for $(x, y) \in [x_0, y_0] \times [x_0 + 1/n, y_0 + 1/n]$. Thus if we let $D_n$ be the closure of the set $[0, 1]^2 - [0, 1/n]^2$, then

$$\sum_{(j,k) \in J_n} \frac{j+k}{(j+1)^3 + (k+1)^3} \leq \int_{D_n} \frac{x+y}{x^3 + y^3} \, dxdy \leq \sum_{(j,k) \in J_n} \frac{j+k+2}{j^3 + k^3}.$$

It is not hard to establish the relation that

$$\sum_{(j,k) \in J_n} \frac{j+k}{(j+1)^3 + (k+1)^3} = \sum_{j,k=1}^{n} \frac{j+k}{j^3 + k^3} + O(1)$$

and that

$$\sum_{(j,k) \in J_n} \frac{j+k+2}{j^3 + k^3} = \sum_{j,k=1}^{n} \frac{j+k}{j^3 + k^3} + O(1).$$

By noting that

\begin{align*} \int_{D_n} \frac{x+y}{x^3 + y^3} \, dxdy &= 2\int_{\frac{1}{n}}^{1} \int_{0}^{y} \frac{x+y}{x^3 + y^3} \, dxdy \\ &= (2 \log n) \int_{0}^{1} \frac{1}{x^2 - x + 1} \, dx \\ &= \frac{4 \pi \log n}{3\sqrt{3}}, \end{align*}

we obtain the asymptotic formula

$$\frac{1}{\log n} \sum_{j,k=1}^{n} \frac{j+k}{j^3 + k^3} = \frac{4 \pi}{3\sqrt{3}} + O\left( \frac{1}{\log n} \right)$$

and hence the answer is $\displaystyle \frac{4 \pi}{3\sqrt{3}}$.

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