MHB Double Sum Challenge: Equate the Limit

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The discussion revolves around evaluating the limit of a double sum given by $$\lim_{n \to \infty} \frac1{n} \sum_{i = 1}^n \sum_{j = 1}^n \frac1{i + j}$$. Participants analyze the equivalence of the double series to a single sum involving the number of pairs $(i, j)$ that sum to a constant $k$. The limit simplifies to a form that includes terms dependent on $n$, leading to a final evaluation of the limit as 2. Corrections to initial solutions are provided, emphasizing the importance of accurate calculations in reaching the correct limit. The challenge highlights various approaches to solving the problem within the mathematical community.
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Equate the limit

$$\lim_{n \to \infty} \frac1{n} \sum_{i = 1}^n \sum_{j = 1}^n \frac1{i + j}$$

Note : This was a challenge from a user in mathstackexchange. From a glance, there should be many ways to do it, so partly I posed this problem to see how the resident analysts in MHB handle it. (although if you think this is solely an analytic problem, you're mistaken ;))
 
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mathbalarka said:
Equate the limit

$$\lim_{n \to \infty} \frac1{n} \sum_{i = 1}^n \sum_{j = 1}^n \frac1{i + j}$$

[sp]The double series is equivalent to ...

$\displaystyle \sum_{i=1}^{n} \sum_{j=1}^{n} \frac{1}{i + j} = \sum_{k=2}^{2 n} \frac{a_{k}}{k}\ (1)$

... where $a_{k}$ is the number of pair of i and j such that i + j = k. It is easy to see that...

$\displaystyle a_{k} = k-1\ \text{if}\ k<2 n, = 1,\ \text{if}\ k=2 n\ (2)$

... so that...

$\displaystyle \lim_{n \rightarrow \infty} \frac{1}{n}\ \sum_{i=1}^{n} \sum_{j=1}^{n} \frac{1}{i + j} = \lim_{n \rightarrow \infty} \frac{1}{n}\ \{ \sum_{k=2}^{2n -1} \frac{k-1}{k} + \frac{1}{2 n} \} = \displaystyle \lim_{n \rightarrow \infty} \frac{2 n - 3}{n} + \frac{1}{2 n^{2}} - \frac{1}{n}\ \sum_{k=2}^{2 n-1} \frac{1}{k} = 2\ (3)$ [/sp]

Kind regards

$\chi$ $\sigma$
 
I am sorry, chisigma, but your solution is incorrect. You are on the right track, however. If you wish to check your solution and repost it, you're more than welcome.
 
mathbalarka said:
Equate the limit

$$\lim_{n \to \infty} \frac1{n} \sum_{i = 1}^n \sum_{j = 1}^n \frac1{i + j}$$

We have

$\displaystyle \lim_{n\to \infty} \frac{1}{n}\sum_{i = 1}^n \sum_{j = 1}^n \frac{1}{i + j}$

$\displaystyle = \lim_{n \to \infty} \frac{1}{n^2} \sum_{i = 1}^n \sum_{j = 1}^n \dfrac{1}{\frac{i}{n} + \frac{j}{n}}$

$\displaystyle = \int_0^1 \int_0^1 \frac{1}{x + y} \, dx\, dy$

$\displaystyle = \int_0^1 (\log(1 + y) - \log(y))\, dy$

$\displaystyle = \int_1^2 \log(y)\, dy - \int_0^1 \log(y)\, dy$

$\displaystyle = (y\log(y) - y)|_{y = 1}^2 - (y\log(y) - y)|_{y = 0}^1$

$\displaystyle = [(2 \log(2) - 2) - (\log(1) - 1)] - [(\log(1) - 1)]$

$\displaystyle = (2\log(2) - 1) + 1$

$\displaystyle = 2\log(2)$.
 
Here's a correction to chisigma's solution.

$$\sum_{i = 1}^n \sum_{j = 1}^n \frac1{i + j} = \sum_{k = 2}^{2n} \frac{c_{k, n}}{k}$$

Where $c_{k, n}$ is the number of partitions of $k$ in two parts, each of size at most $n$. chisigma's flaw in the proof was that he assumed $c_{k, n} = k - 1$, which is true if and only if $k \leq n$. For $k > n$, $c_{k, n} = 2n - k + 1$. Thus,

$$\begin{aligned} \sum_{k = 2}^{2n} \frac{c_{k, n}}{k} = \sum_{k = 2}^{n} \frac{k - 1}{k} + \sum_{k = n + 1}^{2n} \frac{2n - k + 1}{k} \, &= \textcolor{red}{\sum_{k = 2}^n 1} - \textcolor{green}{\sum_{k = 2}^n \frac1{k}} - \textcolor{goldenrod}{\sum_{k = n + 1}^{2n} 1} + \textcolor{blue}{\sum_{k = n+1}^{2n} \frac{2n + 1}{k}} \\ &= \textcolor{red}{(n - 1)} \, - \, \textcolor{green}{\left ( H_n - 1 \right )} \, - \, \textcolor{goldenrod}{n} \, + \, \textcolor{blue}{(2n + 1) \left ( H_{2n} - H_n\right )} \\ &= -H_n + (2n + 1)\left ( H_{2n} - H_n \right ) \end{aligned}$$

Using $H_n \sim \log(n)$ and that $\log(n) = o(n)$

$$\lim_{n \to \infty} \frac1n \sum_{i = 1}^n \sum_{j = 1}^n \frac1{i + j} = \lim_{n \to \infty} \frac{(2n+1)( H_{2n} - H_n ) - H_n}{n} = \lim_{n \to \infty} \left [ \left ( 2 + \frac1n \right) \log(2) - \frac{\log(n)}{n}\right ] = 2 \log(2)$$

What could possibly be a better way to puzzle an analyst than solving this problem number theoretically? :D
 
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