$\lim\limits_{n\to\infty}\frac{n}{n^2+j^2}$

Question

найдем предел вражения

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

запишем, используя методику :

let $S_{n}=\sum_{j=1}^{n^2}\dfrac{n}{n^2+j^2}=\sum_{j=1}^{n^2}\dfrac{1}{1+\left(\dfrac{j}{n}\right)^2}\dfrac{1}{n}$ since $\int_{\dfrac{j}{n}}^{\dfrac{j+1}{n}}\dfrac{dx}{1+x^2}<\dfrac{1}{1+\left(\dfrac{j}{n}\right)^2}\cdot\dfrac{1}{n}<\int_{\dfrac{j-1}{n}}^{\dfrac{j}{n}}\dfrac{dx}{1+x^2}$ so $\int_{\dfrac{1}{n}}^{\dfrac{n^2+1}{n}}\dfrac{dx}{1+x^2}<S_{n}<\int_{0}^{n}\dfrac{dx}{1+x^2}$

и отметим

$\lim_{n\to\infty}\int_{\dfrac{1}{n}}^{\dfrac{n^2+1}{n}}\dfrac{dx}{1+x^2}=\lim_{n\to\infty}\int_{0}^{n}\dfrac{dx}{1+x^2}=\int_{0}^{infty}\dfrac{dx}{1+x^2}=\dfrac{\pi}{2}$ so $\lim_{n\to\infty}\sum_{j=1}^{n^2}\dfrac{n}{n^2+j^2}=\dfrac{\pi}{2}$

Другой способ нахождения предела

$\lim_{n\to\infty}\sum_{j=1}^{n^2}\dfrac{n}{n^2+j^2}=\lim_{n\to\infty}\int_{0}^{n}\dfrac{1}{1+x^2}dx=\dfrac{\pi}{2}$ Но этот метод не совсем правильный! Почему?

Answer 1

Another approach. Since:

$\int_{0}^{+\infty}\sin(ax)e^{-bx}\,dx = \frac{a}{a^2+b^2} \tag{1}$ we have:

$\sum_{j=1}^{n^2}\frac{n}{n^2+j^2} = \int_{0}^{+\infty}\frac{1-e^{-n^2 x}}{e^x-1}\sin(nx)\,dx \tag{2}$ where: $\int_{0}^{+\infty}\frac{\sin(nx)}{e^x-1}\,dx = \text{Im}\int_{0}^{+\infty}\frac{e^{inx}}{e^x-1}\,dx=\sum_{k\geq 1}\frac{n}{n^2+k^2}=\frac{-1+\pi n \coth(\pi n)}{2n}\tag{3}$ by Frullani's theorem and/or the logarithmic derivative of the Weierstrass product for the $\sinh$ function. It is quite trivial that the limit of the RHS of $(3)$ as $n\to +\infty$ is $\frac{\pi}{2}$ , hence it is enough to prove that the contribute given by $\int_{0}^{+\infty}\frac{\sin(nx) e^{-n^2 x}}{e^x-1}\,dx\quad\text{or}\quad\sum_{k>n^2}\frac{n}{n^2+k^2}\tag{4}$ is negligible. But that is quite easy.

Answer 2

Here is a method in the same spirit as what you have done. The main difference is that we explicitly estimate the difference between your sum and $\int_0^n\frac{{\rm d}x}{1+x^2}$ .

Since $f$ is monotonely decreasing we have

$\left|\sum_{j=1}^n \frac{n}{n^2+(j+nm)^2} - \int_{m}^{m+1}\frac{{\rm d}x}{1+x^2}\right| < \frac{1}{n(1 + (m+1)^2)}$

Using this we can estimate the difference between your sum and the integral of $\frac{1}{1+x^2}$ over $[0,n]$ , which can be written $\sum_{m=0}^{n-1}\int_0^1\frac{{\rm d}x}{1+(x+m)^2}$ , as

$\left|\sum_{j=1}^{n^2}\frac{n}{n^2 + j^2} - \int_0^n\frac{{\rm d}x}{1+x^2}\right| \leq \sum_{m=0}^{n-1} \left|\frac{1}{n}\sum_{j=1}^n \frac{1}{1+\left(\frac{j}{n}+m\right)^2} - \int_{0}^{1}\frac{{\rm d}x}{1+(x+m)^2}\right| \\\leq \frac{1}{n}\sum_{m=1}^{n} \frac{1}{1+m^2} \leq \frac{1}{n}\sum_{m=1}^\infty \frac{1}{1+m^2}$

This last sum can be evaluated as $\frac{\pi\coth(\pi)-1}{2}$ , however all we need for this argument is that it's finite. This can be proved for example by comparison to $\sum_{m=1}^\infty \frac{1}{m^2} = \frac{\pi^2}{6}$ . Taking $n\to\infty$ the result follows

$\lim_{n\to\infty}\sum_{j=1}^{n^2}\frac{n}{n^2 + j^2} = \lim_{n\to\infty}\int_0^n\frac{{\rm d}x}{1+x^2} = \frac{\pi}{2}$

Решение задач

Пределы задача. Найти предел $\lim\limits_{n\to\infty}\sum\limits_{j=1}^{n^2}\frac{n}{n^2+j^2}$

$\lim\limits_{n\to\infty}\frac{n}{n^2+j^2}$

Популярные репетиторы:

Решение задач

Пределы задача. Найти предел limn→∞n2∑j=1nn2+j2\lim\limits_{n\to\infty}\sum\limits_{j=1}^{n^2}\frac{n}{n^2+j^2}

limn→∞nn2+j2\lim\limits_{n\to\infty}\frac{n}{n^2+j^2}

Популярные репетиторы:

Запись на занятия

Пределы задача. Найти предел $\lim\limits_{n\to\infty}\sum\limits_{j=1}^{n^2}\frac{n}{n^2+j^2}$

$\lim\limits_{n\to\infty}\frac{n}{n^2+j^2}$