Answer to Question #203241 in Real Analysis for Raj kumar

Question

Show that the series n=1∑∞ x/(1+n²x²) is uniformly convergent in [α,1] for any α>1.

Answer 1

$\sum_{n=1}^{\infty} \frac{x}{1+n^2x^2} ,x\in [\alpha, 1] \text{for any}\alpha< 1.\\ Then,\\ f_n=\frac{x}{1+n^2x^2} \to0\space as\space n\to \infty\\ M_n=sup_{x\in[\alpha, 1]}|f_n-f|\\ =sup_{x\in[\alpha, 1]}|\frac{x}{1+n^2x^2}-0|\\ =sup_{x\in[\alpha, 1]}|\frac{x}{1+n^2x^2}|\\ \text{To get this value, we find the value of x where the given sequence of function is maximum.}\\ f'_n=\frac{1-n^2x^2}{(1+n^2x^2)^2}\\ Now,\\ f'_n=0\implies x=\frac{1}{n}\\ f''_n=\frac{2n^2x(n^2x^2-3)}{(1+n^2x^2)^3}\\ f''_n(\frac{1}{n})<0\\ \implies at \space x=\frac{1}{n}, \space M_n \space \text{attain maximum}.\\ M_n=sup_{x\in[\alpha, 1]}|\frac{x}{1+n^2x^2}|\\ =|\frac{\frac{1}{n} }{1+n^2\frac{1}{n^2}}|\\ =\frac{1}{2n} \to 0\space as \space n\to \infty\\ \text{ Therefore, by Weierstrass M-test the given sequence of function converge. }$