   Chapter 11.6, Problem 44E

Chapter
Section
Textbook Problem

# For which positive integers k is the following series convergent? ∑ n = 1 ∞ ( n ! ) 2 ( k n ) !

To determine

To find: The positive integer k when the series is convergent.

Explanation

Result used: The Ratio Test

“(i) If limn|an+1an|=L<1, then the series n=1an is absolutely convergent (and therefore convergent.)

(ii) If limn|an+1an|=L>1 or limn|an+1an|=, then the series n=1an is divergent.

(ii) If limn|an+1an|=1, the Ratio Test inconclusive; that is, no conclusion can be drawn about the convergence or divergence of n=1an.”a

Calculation:

The given series n=1an=n=1(n!)2(kn)!.

Substitute, n=n+1, then an+1=(n+1)!2(k(n+1))!.

Obtain the limit of |an+1an|.

limn|an+1an|=limn|(n+1)!2(k(n+1))!(n!)2(kn)!|=limn|(n+1)!2(k(n+1))!(kn)!(n!)2|=limn(n+1)2(n!)2[k(n+1)][k(n+1)1][kn+1](kn)!(kn)!(n!)2=limn(n+1)2[k(n+1)][k(n+1)1][kn+1]

If k=1, then limn|an+1an| as follows,

limn|an+1an|=limn(n+1)2[1(n+1)]=limn(n+1)2(n+1)=limnn2</

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