$\sum \frac{1}{n^x}$ for $x \in \left[\frac{4}{\pi},\infty\right)$ converge uniformly.

I'm trying to use the Weierstrass M-Test but having trouble finding an $M_n$.

Any hints?

  • 2
    $\begingroup$ Compare with $\sum 1/n^{4/\pi}$. $\endgroup$ – user296602 Jul 18 '16 at 19:22
  • $\begingroup$ @T. Bongers out of curiosity why not put this as an answer? Thanks. $\endgroup$ – Olivier Oloa Jul 18 '16 at 19:27
  • $\begingroup$ @T.Bongers since $\left|f_n(x)\right|\le M_n$ it suffices for me to check the convergence of $\sum \frac{1}{n^\frac{4}{\pi}}$ to conclude that $\sum \frac{1}{n^x}$ converges uniformly? $\endgroup$ – the_new_guy Jul 18 '16 at 19:35
  • $\begingroup$ @OlivierOloa I don't like giving one line answers, especially when the answer is so straight forward. Feel free to make this an answer if you'd like. $\endgroup$ – user296602 Jul 18 '16 at 19:40
  • $\begingroup$ @the_new_guy Show that $1/n^x \le 1/n^{4\pi}$ for $x$ in this domain, and choose $M_n$ appropriately. $\endgroup$ – user296602 Jul 18 '16 at 19:43

Let $\sum f_n$ denote the series. Note that this series converges uniformly on $[A, \infty)$ for all $A > 1$: we have for all $x \ge A$ and all $n$, $n^x \ge n^A$, hence $1/n^x \le 1/n^A$. Thus,

$$\sup_{x \ge A} f_n(x) \le \frac1{n^A}$$

So you can use Weierstrass' M-test.


The series does not converge uniformly on, say, $(1, \infty)$:

$$R_n(x) = \sum_{k=0}^{\infty} f_k (x) - \sum_{k=0}^n f_k(x) = \sum_{k=n+1}^{\infty} \frac1{k^x} > \sum_{k=n+1}^{2n} \frac1{k^x} > \frac{n}{(2n)^x} = \frac1{2 (2n)^{x-1}}$$

Putting $x_n = 1 + 1/n$, we get:

$$R_n(x_n) \ge \frac1{2(2n)^{1/n}} \to 1/2$$


$$\sup_{x > 1} R_n(x) \not \to 0$$

  • $\begingroup$ Well done. A +1. $\endgroup$ – Mark Viola Jul 18 '16 at 20:27
  • $\begingroup$ Is it true that an appropriate $M_n$ would be $\frac{1}{n}$? I don't completely understand your answer. $\endgroup$ – the_new_guy Jul 19 '16 at 2:12
  • $\begingroup$ @the_new_guy an appropriate $M_n$ would be $1/n^A$ (in your case, $A = 4/\pi$, so the $M_n$ for your question is $1/n^{4/\pi}$). If there's something (besides the note) that you do not understand, please be specific about it. $\endgroup$ – user258700 Jul 19 '16 at 2:16
  • $\begingroup$ This is what I thought, however, I was confused by one of the comments which told me I needed to pick an appropriate $M_n$. $\endgroup$ – the_new_guy Jul 19 '16 at 2:20

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