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

Coroware 2021-11-21 Answered
For which positive integers k is the following series convergent?
\(\displaystyle{\sum_{{{n}={1}}}^{{\infty}}}{\frac{{{n}!}}{{{\left({k}\right)}^{{{n}}}}}}\)

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Expert Answer

Linda Tincher
Answered 2021-11-22 Author has 276 answers

Step 1
Let's use the Ratio Test to check for the series; convergence:
\(\displaystyle{\left|{\frac{{{a}_{{{n}+{1}}}}}{{{a}_{{{n}}}}}}\right|}={\left|{\frac{{{\frac{{{\left({\left({n}+{1}\right)}!\right)}^{{{2}}}}}{{{\left({k}{\left({n}+{1}\right)}\right)}!}}}}}{{{\frac{{{\left({n}!\right)}^{{{2}}}}}{{{\left({k}{n}\right)}!}}}}}}\right|}\)
\(\displaystyle={\frac{{{\frac{{{\left({\left({n}+{1}\right)}!\right)}^{{{2}}}}}{{{\left({k}{\left({n}+{1}\right)}\right)}!}}}}}{{{\frac{{{\left({n}!\right)}^{{{2}}}}}{{{\left({k}{n}\right)}!}}}}}}={\frac{{{\left({\left({n}+{1}\right)}!\right)}^{{{2}}}}}{{{\left({n}!\right)}^{{{2}}}}}}\cdot{\frac{{{\left({k}{n}\right)}!}}{{{\left({k}{\left({n}+{1}\right)}\right)}!}}}\)
\(\displaystyle={\frac{{{\left({n}!\cdot{\left({n}+{1}\right)}\right)}^{{{2}}}}}{{{\left({n}!\right)}^{{{2}}}}}}\cdot{\frac{{{\left({k}{n}\right)}!}}{{{\left({k}{n}+{k}\right)}!}}}\)
\(\displaystyle={\frac{{{\left({n}!\right)}^{{{2}}}\cdot{\left({n}+{1}\right)}^{{{2}}}}}{{{\left({n}!\right)}^{{{2}}}}}}\cdot{\frac{{{\left({k}{n}\right)}!}}{{{\left({k}{n}\right)}!\cdot{\left({k}{n}+{1}\right)}\cdot{\left({k}{n}+{2}\right)}\cdot_{{\cdots}}\cdot{\left({k}{n}+{k}\right)}}}}\)
\(\displaystyle={\frac{{{\left({n}+{1}\right)}^{{{2}}}}}{{{\left({k}{n}+{1}\right)}\cdot{\left({k}{n}+{2}\right)}\cdot_{{\cdots}}\cdot{\left({k}{n}+{k}\right)}}}}\)
Now, let's examine, different values of k to find \(\displaystyle\lim_{{{n}\rightarrow\infty}}{\left|{\frac{{{a}_{{{n}+{1}}}}}{{{a}_{{{n}}}}}}\right|}\). First, for \(\displaystyle{k}={1}\), we have:
\(\displaystyle{\left|{\frac{{{a}_{{{n}+{1}}}}}{{{a}_{{{n}}}}}}\right|}={\frac{{{\left({n}+{1}\right)}^{{{2}}}}}{{{n}+{1}}}}={n}+{1}\rightarrow\infty\),
so the series diverges.
Now let \(\displaystyle{k}={2}\)
\(\displaystyle{\mid}{\frac{{{a}_{{{n}+{1}}}}}{{{a}_{{{n}}}}}}={\frac{{{\left({n}+{1}\right)}^{{{2}}}}}{{{\left({2}{n}+{1}\right)}{\left({2}{n}+{2}\right)}}}}={\frac{{{n}^{{{2}}}+{2}{n}+{1}}}{{{4}{n}^{{{2}}}+{6}{n}+{2}}}}\rightarrow{\frac{{{1}}}{{{4}}}}{<}{1}\)
For \(\displaystyle{k}{>}{2}\), the degree of a polynomial we get in the denominator will be k. The degree of the polinomial in the numeratoe is always 2. Therefore, the limit will always be 0, so the series will be absolutely convergent.
The series will be convergent for all \(\displaystyle{k}\geq{2}\)

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Steacensen69
Answered 2021-11-23 Author has 1470 answers
Step 1
First we need \(\displaystyle{a}_{{{n}}}\rightarrow{0}\). We have
\(\displaystyle{a}_{{{n}}}={\frac{{{\left({n}!\right)}^{{{2}}}}}{{{\left({k}{n}\right)}!}}}\sim{\frac{{{2}\pi{n}{\left({\frac{{{n}}}{{{e}}}}\right)}^{{{2}{n}}}}}{{\sqrt{{{2}\pi{k}{n}}}{\left({\frac{{{k}{n}}}{{{e}}}}\right)}^{{{k}{n}}}}}}\)
\(\displaystyle={\frac{{\sqrt{{{2}\pi{n}}}}}{{\sqrt{{{k}}}}}}{\frac{{{1}}}{{{k}^{{{k}{n}}}}}}{\left({\frac{{{n}}}{{{e}}}}\right)}^{{{\left({2}-{k}\right)}{n}}}\)
Hense, \(\displaystyle{k}\geq{2}\) the series converges by comparison test.
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