Answer to Question #268571 in Calculus for Rakesh

Question

Using Weiestrass M-test, show that the following series converges uniformly.

.

3

1

,

3

1

n x ,x

n 1

3 n 

















 

.

3

1

,

3

1

n x ,x

n 1

3 n 

















 

Answer 1

Question is incomplete.

Let us take an example related to the given problem:

Using Weierstrass’ M-test, show that the series

Infinity

∑ x/n(n+2)^2

n=1

Converges uniformly in [0,k] where k is any given finite positive number.

Solution:

First we give a wording the Weierstrass’ M-test:

Suppose that $\left\{f_n\right\}$ is a sequence of real- or complex-valued functions defined on a set $A$ , and that there is a sequence of positive numbers $\left\{M_n\right\}$ satisfying

$\forall n\ge 1,\quad \forall x\in A:\quad \left|f_n(x)\right|\le M_n\\[0.5cm] \sum\limits_{n=1}^{\infty}M_n<\infty\longrightarrow M_n\, \text{converges}$

Then,

$\sum\limits_{n=1}^{\infty}f_n(x)$

converges absolutely and uniformly on A.

( More information: https://en.wikipedia.org/wiki/Weierstrass_M-test )

In our case,

$\forall x\in[0;k]:\quad\left|f_n(x)\right|=\left|\frac{x}{n(n+2)^2}\right|\le\frac{k}{n(n+2)^2}$

Then, it remains to prove that the series

$\sum\limits_{n=1}^{\infty}\frac{k}{n(n+2)^2}\,\text{converges}$

For this we use the integral test for convergence.

( More information: https://en.wikipedia.org/wiki/Integral_test_for_convergence )

In our case,

$\int_1^\infty\frac{kdx}{x(x+2)^2}=k\cdot\int_1^\infty\frac{dx}{x\cdot x^2(1+2/x)^2}=\\[0.3cm] =\left[\frac{2}{x}=t\to\frac{dx}{x^2}=\frac{dt}{-2};\,x=1\to t=2\,\,\text{and}\,x=\infty\to t=0\right]=\\[0.3cm] =k\cdot\int_2^0\left(\frac{tdt}{-4\cdot (1+t)^2}\right)=\frac{k}{4}\cdot\int_0^2\frac{(t+1)-1dt}{ (1+t)^2}=\\[0.3cm] =\frac{k}{4}\cdot\int_0^2\left(\frac{1}{t+1}-\frac{1}{(t+1)^2}\right)dt=\frac{k}{4}\cdot\left.\left(\ln|t+1|+\frac{1}{t+1}\right)\right|_{t=0}^{t=2}=\\[0.3cm] =\frac{k}{4}\cdot\left(\left(\ln|2+1|+\frac{1}{2+1}\right)-\left(\ln|0+1|+\frac{1}{0+1}\right)\right)=\\[0.3cm] =\frac{k}{4}\cdot\left(\left(\ln3+\frac{1}{3}\right)-\left(\ln|1|+1\right)\right)=\frac{k}{4}\cdot\left(\ln3-\frac{2}{3}\right)<\infty$ $\boxed{\int_1^\infty\frac{kdx}{x(x+2)^2}=\frac{k}{4}\cdot\left(\ln3-\frac{2}{3}\right)\approx k\cdot0.10799}$

Conclusion,

$\text{Since}\quad\sum\limits_{n=1}^{\infty}\frac{k}{n(n+2)^2}\,\text{converges}\to\\[0.3cm] \sum\limits_{n=1}^{\infty}\frac{x}{n(n+2)^2}\,\text{ uniformly converges}$