Interval of convergence of $\sum\limits_{n\geq0} \binom{2n}{n} x^n$

Question

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.

Answer 1

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 $x=1/4$.

Answer 2

$$ \begin{align} \left.\binom{2n}{n}\middle/\binom{2n-2}{n-1}\right. &=\frac{2n(2n-1)}{n^2}\\ &=4-2/n \end{align} $$ Thus, using the ratio test we need to find $x$ so that $$ \limsup_{n\to\infty}(4-2/n)|x|\lt1 $$ and that is when $|x|\lt1/4$.

However, if we use the binomial theorem, we get that $$ \sum_{n=0}^\infty\binom{2n}{n}x^n=(1-4x)^{-1/2} $$ which also indicates a radius of convergence of $\frac14$.

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Endpoints

As zyx points out, $\binom{2n}{n}\ge\frac{4^n}{2n+1}$ since it is the biggest term in row $2n$ of Pascal's Triangle, and the sum of the $2n+1$ terms in that row is $4^n$. Thus, the sum of the series for $x=1/4$ diverges.

Since the ratio of terms for $x=1/4$ is $$ \left.\binom{2n}{n}4^{-n}\middle/\binom{2n-2}{n-2}4^{-n+1}\right.=1-\frac1{2n} $$ and the harmonic series diverges, we get that the terms tend monotonically to $0$$^\ast$. Since the terms for $x=-1/4$ alternate, the series converges for $x=-1/4$.

$^\ast\ $If $x\in(0,1)$, $-\log(1-x)\ge x$, Thus, if $a_n\in[0,1]$ and $\sum\limits_{n=0}^\infty a_n$ diverges, then $\prod\limits_{n=0}^\infty(1-a_n)=0$.