Double Sum Challenge: Equate the Limit

In summary, the limit of the given double series can be rewritten as a single series using the number of pairs of i and j where i + j = k. By finding the limit of this single series, we get the final result of 2.
  • #1
mathbalarka
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0
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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  • #2
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

$\chi$ $\sigma$
 
  • #3
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.
 
  • #4
mathbalarka said:
Equate the limit

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

We have

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

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

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

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

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

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

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

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

$\displaystyle = 2\log(2)$.
 
  • #5
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
 

FAQ: Double Sum Challenge: Equate the Limit

1. What is the Double Sum Challenge: Equate the Limit?

The Double Sum Challenge: Equate the Limit is a mathematical problem that involves finding the limit of a double summation. This means finding the value that a series approaches as the number of terms in the series increases to infinity.

2. How do you solve the Double Sum Challenge: Equate the Limit?

To solve the Double Sum Challenge: Equate the Limit, you need to first simplify the double summation by using algebraic manipulations or known summation formulas. Then, you can apply the limit laws to evaluate the limit of the simplified expression.

3. What are some common techniques used to solve the Double Sum Challenge: Equate the Limit?

Some common techniques used to solve the Double Sum Challenge: Equate the Limit include using the properties of limits, manipulating the double summation to a known form, and using mathematical identities or properties such as telescoping series and geometric series.

4. What are some real-world applications of the Double Sum Challenge: Equate the Limit?

The Double Sum Challenge: Equate the Limit has various real-world applications, such as in physics, engineering, and economics. It can be used to model the behavior of systems that involve the summation of multiple variables, such as the motion of particles, electrical circuits, and financial investments.

5. Are there any tips for solving the Double Sum Challenge: Equate the Limit more efficiently?

One tip for solving the Double Sum Challenge: Equate the Limit more efficiently is to break down the problem into smaller, more manageable steps. This can involve simplifying the expression before evaluating the limit, or breaking the double summation into two separate summations. It is also helpful to have a good understanding of common summation formulas and properties of limits.

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