Let $a_n$ be a sequence of non-negative numbers such that

$$\lim a_n^{\frac{1}{n}}=1$$

Which of the following are correct?

  • $\sum a_n$ converges
  • $\sum a_nx^n$ converges uniformly on $[-\frac{1}{2},\frac{1}{2}]$
  • $\sum a_nx^n$ converges uniformly on $[-1,1]$
  • $\lim \sup\frac{a_{n+1}}{a_n}=1$

My effort:

1.false ,consider $a_n=n$

2.true,The radius of convergence of a power series $\frac{1}{R}=\lim a_n^{\frac{1}{n}}=1\implies R=1$ .Hence the series converges uniformly for compact sets inside $|x|<1$.Hence the series converges uniformly in $|x|\leq 0.5$

3.false ,putting $x=1$ the same as in case 1.

4.I am unable to prove this fact.How to solve this .


First of all, way to go for your efforts. As far as I can see, your answers to the first three questions are correct. To refute the last one consider the sequence


Observe that


  • $\begingroup$ in order to show that $a_n^{\frac{1}{n}}=1$ should i do like this:::. if $n=$ odd then $a_n^{\frac{1}{n}}=1$ and if $n=$ even then $a_n=1\forall n$ hence $a_n^{\frac{1}{n}}=1$ $\endgroup$ – Learnmore Aug 10 '16 at 17:26
  • $\begingroup$ is it okay ;please comment $\endgroup$ – Learnmore Aug 10 '16 at 17:28
  • $\begingroup$ Yes, it is right. If you can split a sequence in a finite set of pairwise disjoint subsequences such that each such subsequence converges to the same limit, then the whole sequence converges to the same common limit. $\endgroup$ – DonAntonio Aug 10 '16 at 17:48
  • $\begingroup$ If you don't mind,can I ask you for a outline of the proof ;I have heard of the result that if every subsequence converges to the same limit then the original sequence converges.But this result is new $\endgroup$ – Learnmore Aug 11 '16 at 2:37
  • $\begingroup$ @S.Bandopadhaya If $n$ is odd, $a_n^{1/n}\ne1$. $\endgroup$ – Did Aug 11 '16 at 13:39

Your answer to 2) is wrong. $R=1$ implies the power series converges absolutely for $|x|<1,$ but not uniformly in that range. Example: $\sum x^n.$ However, the power series will converge uniformly in $[-a,a]$ for all $a\in [0,1).$

  • $\begingroup$ But the example you give does converge uniformly on $\;[-1/2,\,1/2]\;$ , which is what is being asked in (2). $\endgroup$ – DonAntonio Aug 10 '16 at 17:02
  • $\begingroup$ But the reason given by the OP was that the series converges uniformly on $(-1,1).$ That reasoning is incorrect. $\endgroup$ – zhw. Aug 10 '16 at 17:11
  • 1
    $\begingroup$ I think that reasoning is correct: if the power series $\;\sum a_n(z-z_0)^n\;$ has convergence radius $\;R\;$, then for any $\;z\,,\,\,|z-z_0|<R\;$ the convergence is absolute, and it is uniform on any compact subset contained in the circle (interval, in the real case) of convergence. $\endgroup$ – DonAntonio Aug 10 '16 at 17:15
  • $\begingroup$ Did you even read the OP's answer 2)? $\endgroup$ – zhw. Aug 10 '16 at 17:18
  • $\begingroup$ @S. I think I can see zhw's point now and he's right: you wrote at the end of your reasoning in point (2) that the series converges unif. for $\;|z|<1\;$ . This is false. $\endgroup$ – DonAntonio Aug 10 '16 at 17:19

For the last one , a general result can help you : $$\lim\sup\left|\frac{a_{n+1}}{{a_n}}\right|\ge \lim\sup|a_n|^{1/n}$$ $$ \lim\inf|a_n|^{1/n}\ge\lim\inf \left|\frac{a_{n+1}}{{a_n}}\right|$$


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