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Calculus Level 4

$\sum_{n=0}^{\infty} \binom{2n}{n}x^{n}$

If the above sum is a real number, find $\displaystyle \sup x - \inf x$ , where $\sup$ is the supremum, $\inf$ is the infimum, and $x$ is real.

The answer is 0.5.

1 solution

Jake Lai
May 13, 2015

The ratio test shows that the sum converges for $|x| < 0.25$ :

$\lim_{n \rightarrow \infty} \frac{\binom{2(n+1)}{n+1}x^{n+1}}{\binom{2n}{n}x^{n}} = \lim_{n \rightarrow \infty} \frac{(2n+2)(2n+1)x}{(n+1)^{2}} = 4x$

As such, the supremum of $x$ (but not the maximum) is $0.25$ ; likewise, the infimum (but not minimum) is $-0.25$ . Hence,

$\sup x - \inf x = 0.25-(-0.25) = \boxed{0.5}$

Moderator note:

Nice work! Bonus question: What would the answer be if I replace $\dbinom{2n}{n}$ by $\dbinom{mn}{n \ n \ n \ldots n }$ for some positive integer $m$ (note that there's $m$ $n$ 's at the bottom of the multinomial coefficient)?

I'm going to assume that's the multinomial coefficient defined by

$\binom{n}{k_{1}, k_{2}, \ldots , k_{m}} = \frac{n!}{k_{1}!k_{2}! \ldots k_{m}!}$

In your replacement, we have

$\binom{mn}{n, n, \ldots , n} = \frac{(mn)!}{n!n! \ldots n!} = \frac{(mn)!}{n!^{m}}$

By a similar argument as above, using the ratio test shows that

$\lim_{n \rightarrow \infty} \frac{(m(n+1))!}{(n+1)!^{m}} / \frac{(mn)!}{n!^{m}} = \lim_{n \rightarrow \infty} \frac{(mn+1)(mn+2) \ldots (mn+m)}{(n+1)^{m}} = m^{m}$

so we require $|x| < \frac{1}{m^{m}}$ .

Jake Lai - 6 years, 1 month ago

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The answer is 0.5.

1 solution

Moderator note:

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