Show that the radius of convergence of the power series is at least 1 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! 
 A: 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.
A: $|a_i|$ 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).
