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# HELP NEED MATH GENIUS, What is k?

For which positive integers k is the following series convergent?

Sum symbol from n=1 to infinity

(n!)^2/(kn)!

k >= ?!!?

### 1 Answer

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- kbLv 77 years agoFavorite Answer
If k = 1, then this series diverges by the nth term test, because lim(n→∞) n! = ∞.

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Assuming k > 1, now we use the Ratio Test:

r = lim(n→∞) [((n+1)!)^2 / (k(n+1))!] / [(n!)^2 / (kn)!]

..= lim(n→∞) [((n+1)!)^2 / (kn+k)!] * [(kn)! / (n!)^2]

..= lim(n→∞) (n+1)^2 / [(kn+k)(kn+k-1)...(kn-1)].

If k = 2, then r = lim(n→∞) (n+1)^2 / [(2n+2)(2n+1)] = 1/4 < 1.

So, the series converges.

If k > 2, then r = 0 < 1 (since the degree of the denominator is greater that of the numerator).

So, the series again converges.

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Hence, the series converges precisely when k ≥ 2.

I hope this helps!

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