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One problem saying:

Is $\sum_{n=1}^{\infty}1/(x^{2}+n^{2})$ uniform convergence?

So I solved it using Weierstrass M test since $\mathbb R $ is complete and $g_k\leq 1/k^2$ and the series consisting of $1/k^2$ is convergent by P-series test.

But what I want to know is the following problem:

Is $\sum_{n=1}^{\infty}x^{n}/n^{2}$ uniformly convergent on $x\in [0,1]$?

To solve it, I tried to use Cauchy Criterion.

So, I tried to find, for given $\epsilon$, certain $N$ for that condition.

For $k\geq N$

$\|g_{k}(x)+\cdots+g_{k+p}(x)\|=\|\frac{x^{k}}{k^{2}}+\cdots+\frac{x^{k+p}}{(k+p)^{2}}\|\leq\|\frac{1}{k^{2}}+\cdots+\frac{1}{k^{2}}\|=\frac{p}{k^{2}}.$

How can I find such $N$?

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  • $\begingroup$ No, it cannot be convergent, as $\sum_{n=1}^\infty x^n/n^2$ has convergence radius $1$ $\endgroup$ – Ákos Somogyi Nov 16 '15 at 9:34
  • $\begingroup$ Your second series is convergent in $[-1,1]$ by the $M$-test. $\endgroup$ – Martín-Blas Pérez Pinilla Nov 16 '15 at 9:34
  • $\begingroup$ Ah! It's all my fault. problem gives me "Only $x\in[0,1]$" condition $\endgroup$ – Darae-Uri Nov 16 '15 at 9:36
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If $n \geq 1$, then $$ x^{n}/n^{2} \leq 1/n^{2} $$ for all $x \in [0,1]$; by comparison test the uniform convergence follows.

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  • $\begingroup$ Sorry for confusion. I fixed it. Thank you for your answer. $\endgroup$ – Darae-Uri Nov 16 '15 at 9:41
  • $\begingroup$ Then what value should I take as $N$? If I choose $N=1/\sqrt \epsilon$ then for $n\geq N$, $||g_k (x)+\cdots +g_{k+p} (x)||\leq||1/n^2 +\cdots +1/n^2 ||=p/\epsilon. $ $\endgroup$ – Darae-Uri Nov 16 '15 at 9:56
  • $\begingroup$ @Darae-Uri I was suggesting a quicker and to me more elegant way to see the uniform convergence; Cauchy's stuff can make things messy. :) $\endgroup$ – Megadeth Nov 16 '15 at 10:12
  • $\begingroup$ Ah! I see. Thank you for your comment. :D $\endgroup$ – Darae-Uri Nov 16 '15 at 10:13
  • $\begingroup$ @Darae-Uri Sorry for the careless version of my previous answer. $\endgroup$ – Megadeth Nov 16 '15 at 10:17

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