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

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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Linda Tincher

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 $$\displaystyle{k}\geq{2}$$

###### Have a similar question?
Steacensen69
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.