Answer to Question #216789 in Real Analysis for subodh gupta

Question

Test the uniform convergence of the series

Σ∞

𝑛=1 x[𝑛/(1+𝑛^2𝑥^2)−(𝑛+1)/(1+(𝑛+1)^2𝑥^2)]

Answer 1

$Σ^∞_{𝑛=1} x[\frac{𝑛}{(1+𝑛^2𝑥^2)−\frac{(𝑛+1)}{(1+(𝑛+1)^2𝑥^2)}}]$

We can reduce the series to a simpler form to work with as shown below

$Σ^∞_{𝑛=1} [\frac{x}{(1+𝑛𝑥)((n-1)x+1)}]$

The above series is uniformly convergent on any finite interval [a; b], when $0 < a < b$ . But the series is only point wise convergent but not uniformly convergent on [a, b]

Here $U_n(x)= [\frac{x}{(1+𝑛𝑥)((n-1)x+1)}] = \frac{1}{(n-1)x+1}-\frac{1}{nx+1}$

Therefore $s_n(x) = u_1(x) + u_2(x) + u_3(x) + · · · + u_n(x)$

$s_n(x)= (1-\frac{1}{x+1})+(\frac{1}{x+1}+\frac{1}{2x+1})+(\frac{1}{2x+1}+\frac{1}{3x+1})+...+[\frac{1}{(n-1)x+1}-\frac{1}{nx+1}]\\ s_n(x)= 1-\frac{1}{nx+1}\\ \therefore s(x)=Lim_{n \to \infin} s_n(x)= x = \begin{cases} 1 &\text{if } x\not=0 \\ 0 &\text{if } x=0 \end{cases}$

Therefore the sum function s(x) is discontinuous on [a, b], and therefore the convergence is not uniform

on [a, b], it is only point wise. When $x \not= 0$ , let $\epsilon > 0$ be given

$|s_n(x)-s(x)|=|1-\frac{1}{nx+1}-1|=|\frac{1}{nx+1}| < \epsilon$ ; whenever $n > \frac{1}{x}(\frac{1}{\epsilon}-1)$

But $\frac{1}{x}(\frac{1}{\epsilon}-1)$ decreases with x. Hence if we take  $N= [\max\limits_{x \epsilon[a,b] } \begin{cases} \end{cases} \frac{1}{x}(\frac{1}{\epsilon}-1)] +1\epsilon N$

which is independent of x, then we obtain $|s_n(x)-s(x)|< \epsilon$ whenever n> N whenever n > N for all $x \epsilon [a, b]$ i.e. the series converges uniformly to s(x) = 1 on [a,b].