Convergence (1)

Relevant wiki: Convergence - Ratio Test

The question asks about the convergence of $\displaystyle\sum_{n=1}^{\infty}{\dfrac{m^n((500n)!)^2}{(1000n)!}}$ so we can proceed by the ratio test. $\lim_{n \to \infty}{\dfrac{m^{n+1}((500n+500)!)^2(1000n)!}{m^n((500n)!)^2(1000n+1000)!}} \\ = m\lim_{n \to \infty}{\dfrac{((500n+1)\times(500n+2)\times(500n+3)\times\cdots\times(500n+500))^2}{(1000n+1)\times(1000n+2)\times(1000n+3)\times\cdots\times(1000n+1000)}}$

Since each factor in the numerator is repeated once, we note that there are $1000$ factors in the numerator and the denominator of the fraction. Since $\displaystyle\lim_{n \to \infty}{\large{\dfrac{\prod_{k=1}^{500}{(500n+k)}}{\prod_{k=1}^{500}{(1000n+k)}}}} = \dfrac{1}{2^{500}}$ we have $\dfrac{m}{2^{500}}\lim_{n \to \infty}{\large{\dfrac{\prod_{k=1}^{500}{(500n+k)}}{\prod_{k=501}^{1000}{(1000n+k)}}}}$

Once again, since $\displaystyle\lim_{n \to \infty}{\large{\dfrac{\prod_{k=1}^{500}{(500n+k)}}{\prod_{k=501}^{1000}{(1000n+k)}}}} = \dfrac{1}{2^{500}}$ we have $\dfrac{m}{2^{500}}\lim_{n \to \infty}{\large{\dfrac{\prod_{k=1}^{500}{(500n+k)}}{\prod_{k=501}^{1000}{(1000n+k)}}}} = \dfrac{m}{2^{500}}\times\dfrac{1}{2^{500}} = \dfrac{m}{2^{1000}}$ Since we used a ratio test to determine the convergence of the series, we conclude that since $\dfrac{|m|}{2^{1000}}$ must be less than $1$ , then $m < 2^{1000}$ .

Hence, $s+r = 1000+2 = \boxed{1002}$ .

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Furthermore, if we replace $500$ with $k$ and $1000$ with $2k$ we see that a series of the form $\displaystyle{\sum_{n=1}^{\infty}{\dfrac{m^n((kn)!)^2}{(2kn)!}}}$ is convergent if $|m| < 2^{2k}$ .

Relevant wiki: Convergence - Ratio Test

$\begin{aligned} \sum_{n=0}^\infty a_n & = \sum_{n=0}^\infty \frac {m^n((500n)!)^2}{(1000n)!} & \small \color{#3D99F6} \text{Stirling's formula: }n! \approx \sqrt {2\pi} n^{n+1/2} e^{-n} \\ & = \sum_{n=0}^\infty \frac {m^n \sqrt {2\pi}(500n)^{1000n+1}}{(1000n)^{1000n+1/2}} \end{aligned}$

Consider the ratio text, for the sum to converge, we have $\displaystyle L = \lim_{n \to \infty} \left| \frac {a_{n+1}}{a_n} \right| < 1$ . Therefore,

$\begin{aligned} L & = \lim_{n \to \infty} \frac {|m|^{n+1} \sqrt {2\pi}(500(n+1))^{1000(n+1)+1}}{(1000(n+1))^{1000(n+1)+1/2}} \cdot \frac {(1000n)^{1000n+1/2}} {|m|^n \sqrt {2\pi}(500n)^{1000n+1}} \\ & = \frac {|m|(500n)^{1000(n+1)+1}{\color{#3D99F6}\left(1+\frac 1n\right)^{1000(n+1)+1}}(1000n)^{1000n+1/2}}{(1000n)^{1000(n+1)+1/2}{\color{#3D99F6}\left(1+\frac 1n\right)^{1000(n+1)+1/2}}(500n)^{1000n+1}} \\ & = \frac {|m| (500n)^{1000}\color{#3D99F6}e^{1000}}{(1000n)^{1000}\color{#3D99F6}e^{1000}} \\ & = \frac {|m|}{2^{1000}} \end{aligned}$

For $L < 1 \implies \dfrac {|m|}{2^{1000}} < 1 \implies |m| < 2^{1000}$ . Therefore, $s+r = 2+1000 = \boxed{1002}$ .

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References:

• Stirling's formula
• Ratio test