<s>[INST] Interval of convergence of $\sum\limits_{n\geq0} \binom{2n}{n} x^n$ We consider the power series $\displaystyle{\sum_{n\geq0} {2n \choose n} x^n}$. By Ratio Test, the radius of convergence is easily shown to be $R=\frac{1}{4}$.
For $x=\frac{1}{4}$, Stirling equivalent and Ratio Test imply that the series is divergent. For $x=\frac{-1}{4}$, Stirling equivalent and Alternating Series Test can be used to show the convergence of the series. Thus the interval of convergence is $\left[\frac{-1}{4},\frac{1}{4}\right)$.
My question: is there a (preferably simple ^^) method to determine the interval of convergence without using equivalents? My students don't know about equivalents. 
 [/INST] Let $a_n=\frac{1}{4^n}\binom{2n}{n}$. Then
\begin{eqnarray}
a_n&=&\frac{(2n)\cdot(2n-1)\cdot\dots\cdot 2 \cdot 1}{(2n)^2 \cdot (2n-2)^2 \cdot \dots \cdot 2} \\ 
&=&\frac{(2n-1) \cdot (2n-3) \cdot \dots \cdot 3 \cdot 1}{(2n) \cdot (2n-2) \cdot \dots \cdot 4 \cdot 2} \\
&=& \prod_{k=1}^n\left(1-\frac{1}{2k}\right)\\
&<& \prod_{k=1}^n \sqrt{\left(1-\frac{1}{2k+1}\right)\left(1-\frac{1}{2k}\right)}\\
&=&\sqrt{\frac{(2n)!}{(2n+1)!}}\\
&=&\sqrt{\frac{1}{2n+1}} \, .
\end{eqnarray}
So $a_n$ tends to $0$ as $n \to \infty$; as $a_n$ is clearly monotone, the alternating series test proves convergence for $x=-1/4$.
You can make a similar comparison to show that $a_n>\sqrt{\frac{1}{4n}}$, which proves divergence for $x=1/4$.
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