# Which of the following facts are true of a sequence satisfying $\lim a_n^{\frac{1}{n}}=1$?

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

$$\{a_n\}=\{1,1,2,1,3,1,4,1,5,1,6,1,....\}=\begin{cases}k,&n=2k-1\\{}\\1,&\text{otherwise}\end{cases}$$

Observe that

$$\left\{\frac{a_{n+1}}{a_n}\right\}=\left\{1,2,\frac12,3,\frac13,4,...\right\}$$

• 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 $a_n^{\frac{1}{n}}=1$ – Learnmore Aug 10 '16 at 17:26
• is it okay ;please comment – Learnmore Aug 10 '16 at 17:28
• 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. – DonAntonio Aug 10 '16 at 17:48
• 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 – Learnmore Aug 11 '16 at 2:37
• @S.Bandopadhaya If $n$ is odd, $a_n^{1/n}\ne1$. – 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 $a\in [0,1).$

• But the example you give does converge uniformly on $\;[-1/2,\,1/2]\;$ , which is what is being asked in (2). – DonAntonio Aug 10 '16 at 17:02
• But the reason given by the OP was that the series converges uniformly on $(-1,1).$ That reasoning is incorrect. – zhw. Aug 10 '16 at 17:11
• 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. – DonAntonio Aug 10 '16 at 17:15
• Did you even read the OP's answer 2)? – zhw. Aug 10 '16 at 17:18
• @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. – DonAntonio Aug 10 '16 at 17:19

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