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If the coefficients ${a_i}$ of a power series $\displaystyle\sum_{i=0}^{\infty}a_{i}x^{i}$ form a bounded sequence show that the radius of convergence of the power series is at least $1$

How to solve this? Please clearly show the proof of this question. Thank you so much!

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    $\begingroup$ Can you find a lower bound for $1/\limsup\limits_{n\rightarrow\infty} \root n\of{ |a_n|}$? $\endgroup$ – David Mitra Feb 23 '13 at 13:11
  • $\begingroup$ @DavidMitra Ok.please look at answer2. And I have a question. Now you show at least 1. Well, can we say that equal to one. That's, does we say R=1 directly without any extra proof? $\endgroup$ – user315 Feb 23 '13 at 14:40
  • $\begingroup$ sbr's method is the quicker approach. The second answer has typos near the end (the inequalities are reversed). One has $\limsup\limits_{n\rightarrow\infty}\root n\of{|a_n|}\le 1$. From this, the best you can say is that the radius of convergence is at least $1$. $\endgroup$ – David Mitra Feb 23 '13 at 14:46
  • $\begingroup$ Hmm ok @DavidMitra please you can explain how to show that R=1? Please help! $\endgroup$ – user315 Feb 23 '13 at 14:51
  • $\begingroup$ $R$ is not necessarily equal to $1$. As I said, the best you can get is $R\ge1$. $\endgroup$ – David Mitra Feb 23 '13 at 14:55
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Suppose that the sequence $\{|a_i|\}$ is bounded above by $M\geq0$ so we have $$\sum|a_i x^i|\leq M\sum |x^i|$$ and the radius of convergence of the last serie is $1$, then you can conclude.

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$|a_i|$ is bounded which implies $\exists M\geq0 $ such that $|a_i|\le M \forall i\in N$

Now if $M<1$ then we have $|a_i|^{\frac{1}{n}}\le M^{\frac{1}{n}}<1$ So $\lim\sup_{n\to \infty}|a_i|^{\frac{1}{n}}<1$

Else if $M\ge 1$ then let $M=1+k,k\ge 0$

We know,

$(1+k/n)^{n}\ge (1+k)$

$\Rightarrow (1+k/n)\ge (1+k)^{\frac{1}{n}}$

So we have ,

$|a_i|^{\frac{1}{n}}\le M^{\frac{1}{n}}\le (1+k)^{\frac{1}{n}}\le (1+k/n)$

$\Rightarrow \lim\sup_{n\to \infty}|a_i|^{\frac{1}{n}}\le \lim\sup_{n\to \infty}(1+k/n)=1$(As the limit of the sequence $(1+k/n)$ exists and is equal to 1 so $\lim\sup_{n\to \infty}(1+k/n)=1$)

So we have $\lim\sup_{n\to \infty}|a_i|^{\frac{1}{n}}\le 1$

$\Rightarrow \displaystyle \frac{1}{\lim\sup_{n\to \infty}|a_i|^{\frac{1}{n}}}\ge 1$

$\Rightarrow R\ge 1$(Here R is the radius of convergence).

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  • $\begingroup$ Ok. Thank you! And I have a question. Now you show at least 1. Well, can we say that equal to one. That's, does we say R=1 directly without any extra poof? $\endgroup$ – user315 Feb 23 '13 at 14:30
  • $\begingroup$ @Bs11 sorry for the typo that was there in the last two lines. Now I have corrected it. As an answer to you question it is not necessary for R to be 1 it can be $>1$ also. $\endgroup$ – Abhra Abir Kundu Feb 23 '13 at 15:26
  • $\begingroup$ Ok. No problem! I have noticed this mistake. Thank you $\endgroup$ – user315 Feb 23 '13 at 15:27
  • $\begingroup$ @Bs11 if you take $a_{n}=2^{-n}$ then the radius of convergence is 2(which is not equal to 1. $\endgroup$ – Abhra Abir Kundu Feb 23 '13 at 15:30

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