Answer to Question #216789 in Real Analysis for subodh gupta

Question #216789

Test the uniform convergence of the series

Σ∞

𝑛=1 x[𝑛/(1+𝑛^2π‘₯^2)βˆ’(𝑛+1)/(1+(𝑛+1)^2π‘₯^2)]


1
Expert's answer
2021-07-16T02:41:38-0400

Σ𝑛=1∞x[𝑛(1+𝑛2π‘₯2)βˆ’(𝑛+1)(1+(𝑛+1)2π‘₯2)]Ξ£^∞_{𝑛=1} x[\frac{𝑛}{(1+𝑛^2π‘₯^2)βˆ’\frac{(𝑛+1)}{(1+(𝑛+1)^2π‘₯^2)}}]

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

Σ𝑛=1∞[x(1+𝑛π‘₯)((nβˆ’1)x+1)]Ξ£^∞_{𝑛=1} [\frac{x}{(1+𝑛π‘₯)((n-1)x+1)}]

The above series is uniformly convergent on any finite interval [a; b], when 0<a<b0 < a < b . But the series is only point wise convergent but not uniformly convergent on [a, b]

Here Un(x)=[x(1+𝑛π‘₯)((nβˆ’1)x+1)]=1(nβˆ’1)x+1βˆ’1nx+1U_n(x)= [\frac{x}{(1+𝑛π‘₯)((n-1)x+1)}] = \frac{1}{(n-1)x+1}-\frac{1}{nx+1}

Therefore sn(x)=u1(x)+u2(x)+u3(x)+β‹…β‹…β‹…+un(x)s_n(x) = u_1(x) + u_2(x) + u_3(x) + Β· Β· Β· + u_n(x)

sn(x)=(1βˆ’1x+1)+(1x+1+12x+1)+(12x+1+13x+1)+...+[1(nβˆ’1)x+1βˆ’1nx+1]sn(x)=1βˆ’1nx+1∴s(x)=Limnβ†’βˆžsn(x)=x={1if x=ΜΈ00if x=0s_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=ΜΈ0x \not= 0 , let Ο΅>0\epsilon > 0 be given

∣sn(x)βˆ’s(x)∣=∣1βˆ’1nx+1βˆ’1∣=∣1nx+1∣<Ο΅|s_n(x)-s(x)|=|1-\frac{1}{nx+1}-1|=|\frac{1}{nx+1}| < \epsilon ; whenever n>1x(1Ο΅βˆ’1)n > \frac{1}{x}(\frac{1}{\epsilon}-1)

But 1x(1Ο΅βˆ’1)\frac{1}{x}(\frac{1}{\epsilon}-1) decreases with x. Hence if we take  N=[max⁑xΟ΅[a,b]{1x(1Ο΅βˆ’1)]+1Ο΅NN= [\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 ∣sn(x)βˆ’s(x)∣<Ο΅|s_n(x)-s(x)|< \epsilon whenever n> N whenever n > N for all xΟ΅[a,b]x \epsilon [a, b] i.e. the series converges uniformly to s(x) = 1 on [a,b].


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