Question #288897

Show that the series , ∞Σ n=1 (x/1+n^2x^2) is uniformly convergent in [ k,1] for any k>0.

1
Expert's answer
2022-01-20T08:17:06-0500

If x[k,1]x\in[k,1] , 0<k<10<k<1 , then x1+n2x2xn2x2=1n2x1n2k\frac{x}{1+n^2x^2}\leq \frac{x}{n^2x^2}=\frac{1}{n^2x}\leq \frac{1}{n^2 k}.

The expression on the right hand does not depend on x and the series n=11n2k\sum\limits_{n=1}^{\infty}\frac{1}{n^2 k} is convergent, therefore the initial series is uniformly convergent on the segment [k,1].


In order to prove that the series is uniformly convergent for any k>0k>0 , it is sufficient to show that for any NNN\in\mathbb{N} there exists x[0,1]x\in[0,1] , q>p>Nq> p> N such that n=pqx1+n2x21/5\sum\limits_{n=p}^{q}\frac{x}{1+n^2x^2}\geq 1/5 .

Put x=1/N, p=N+1, q=2N. Then nx2nx\leq 2 for all n=p,p+1,,qn=p, p+1, \dots, q and

n=pqx1+n2x2n=pqx1+22=n=N+12N1/N5=1/5\sum\limits_{n=p}^{q}\frac{x}{1+n^2x^2}\geq \sum\limits_{n=p}^{q}\frac{x}{1+2^2}=\sum\limits_{n=N+1}^{2N}\frac{1/N}{5}=1/5

Therefore, the series n=1x1+n2x2\sum\limits_{n=1}^{\infty}\frac{x}{1+n^2x^2} is uniformly convergent for any k>0k>0 .



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