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Find the smallest value of $b$ that makes the following statement true:

$$\text{If} \quad 0\leq a<b, \text{then the series} \ \sum\limits_{n=1}^{\infty}\dfrac{(n!)^2a^n}{(2n)!}\ \text{converges}$$

A) $1$ $\ $ B) $2\log 2$ $\ $ C) $2$ $\ $D) $\sqrt{2}$$\ $ E) $4$

My solution: I have applied Cauchy theorem to above series and derived that the radius of convergence is $4$. Indeed, $$\dfrac{1}{R}=\lim \limits_{n \to \infty} \dfrac{a_{n+1}}{a_n}=\lim \limits_{n\to \infty} \dfrac{((n+1)!)^2}{(2n+2)!}\cdot \dfrac{(2n)!}{(n!)^2}=\dfrac{1}{4}$$ Hnce, $R=4$. So above power series converges for any $a$ such that $|a|<4$.

But the purpose of the problem is to find the smallest value of $b$. Among five options the smallest is option A) because power series converges for any $a \in [0, 1)$.

But my answer is incorrect. Can anyone explain it in detail please?

RFZ
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1 Answers1

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Technically, you are correct. Among the given answer choices, the smallest value that makes the statement true corresponds to choice (A). The problem is poorly posed; the intent was to find the supremum of all $a$ such that $$\sum_{n=1}^\infty \frac{(n!)^2 a^n}{(2n)!}$$ converges; that is to say, the smallest value $b$ for which all $a$ satisfying $0 \le a < b$ results in the series converging. This was not, unfortunately, how the actual question was written, leading to a technical ambiguity. Under your technically correct reasoning, any arbitrarily small positive $b \le 4$ is trivially permissible, but this makes the question "uninteresting."

heropup
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