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Consider the series $$\sum\limits_{n=1}^\infty\frac{(2n)!}{a^n(n!)^2}$$ with $a > 0$. Determine if the series converges for:

i) a > 4

ii) 0 < a < 4

iii) a = 4

For i) and ii), I will use the ratio test, $$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\frac{4}{a}$$

So, if $a > 4$, the series converge and, if $0 < a < 4$, the series do not.

iii) If $a = 4$, I am a bit stuck. Can someone please give me some tips? Thank you

darkchampionz
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1 Answers1

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Your coefficients are $${2n\choose n}\frac{1}{4^n}\ge \frac{1}{n+1}$$ where ${2n\choose n}$ are central binomial coefficients, and I used the bound ${2n\choose n}\ge \frac{4^n}{n+1}$. Hence by comparison with the harmonic series, this diverges.

vadim123
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