For which positive integers k is the following series convergent? \sum_{n

Coroware

Coroware

Answered question

2021-11-21

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

Answer & Explanation

Linda Tincher

Linda Tincher

Beginner2021-11-22Added 14 answers

Step 1
Let's use the Ratio Test to check for the series; convergence:
|an+1an|=|((n+1)!)2(k(n+1))!(n!)2(kn)!|
=((n+1)!)2(k(n+1))!(n!)2(kn)!=((n+1)!)2(n!)2(kn)!(k(n+1))!
=(n!(n+1))2(n!)2(kn)!(kn+k)!
=(n!)2(n+1)2(n!)2(kn)!(kn)!(kn+1)(kn+2)(kn+k)
=(n+1)2(kn+1)(kn+2)(kn+k)
Now, let's examine, different values of k to find limn|an+1an|. First, for k=1, we have:
|an+1an|=(n+1)2n+1=n+1,
so the series diverges.
Now let k=2
an+1an=(n+1)2(2n+1)(2n+2)=n2+2n+14n2+6n+214<1
For k>2, the degree of a polynomial we get in the denominator will be k. The d

Steacensen69

Steacensen69

Beginner2021-11-23Added 15 answers

Step 1
First we need an0. We have
an=(n!)2(kn)!2πn(ne)2n2πkn(kne)kn
=2πnk1kkn(ne)(2k)n
Hense, k2 the series converges by comparison test.

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