Answer to Question #111258 in Calculus for Gaurav

Question

Use Weierstrass’ M-Test to prove that the series ∑(x^nXn^2)
∞
n=1

n converges uniformly in x
the interval .
5
1
,0



Answer 1

Let's substitute the edges of the interval to the initial series $\sum x^2n^2.$

1. $x = 0$ :

The result is 0

2. $x = 1/5$ :

The result is $\sum \dfrac{n^2}{5^n}.$

It is obvious, that for any intermediate value of x from this inteval:

$\sum x^nn^2 \le \sum\frac{n^2}{5^n}.$

The series $\sum\frac{n^2}{5^n}$ converges according to the d'Alembert's ratio test. To prove it let's find the limit:

$\lim_{n \to \infty} |\dfrac{a_{n+1}}{a_n}| = \lim_{n \to \infty} \dfrac{(n+1)^2 5^n}{5^{n+1}n^2} =\\=\dfrac15 \lim_{n \to \infty} \dfrac{n^2+2n+1}{n^2} = \dfrac15 \lim_{n \to \infty} (1 + \dfrac2n + \dfrac{1}{n^2}) = \dfrac15$ .

As far as $\frac15<1$ , the series $\sum\frac{n^2}{5^n}$ converges according to the d'Alembert's ratio test.

Thus, by the Weierstrass’ M-Test the initial series converges uniformly on the interval [0,1/5].

QED