Convergence using the ratio test

Question

I am trying to determine the smallest value for natural number $k$ such that $$\sum\limits_{n=1}^\infty\frac{(n!)^{k-1}(27n)^n}{(kn)!e^n}$$ converges. I attempted to solve using the ratio test:

$$\text{If} \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=L<1, \text{then the series $\sum\limits_{n=1}^\infty a_n$ is absolutely convergent.}$$

Using $a_n = \frac{(n!)^{k-1}(27n)^n}{(kn)!e^n}$ and $a_{n+1} = \frac{((n+1)!)^{k-1}(27(n+1))^{n+1}}{(k(n+1))!e^{n+1}}$, I tried to simplify the following ratio like this:

$$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|= \lim_{n\to\infty}\left|\frac{\frac{((n+1)!)^{k-1}(27(n+1))^{n+1}}{(k(n+1))!e^{n+1}}}{\frac{(n!)^{k-1}(27n)^n}{(kn)!e^n}}\right|$$

$$= \lim_{n\to\infty}\left|\frac{\frac{(n+1)^{k-1}(n!)^{k-1}(27)^{n+1}(n+1)^{n+1}}{(kn+k)(kn+(k-1))\cdots(kn+1)(kn)!e^{n+1}}}{\frac{(n!)^{k-1}(27^n)n^n}{(kn)!e^n}}\right|$$

$$= \frac{27^{n+1}e^n}{27^ne^{n+1}}\lim_{n\to\infty}\left|\frac{\frac{(n+1)^{k-1}(n+1)^{n+1}}{(kn+k)(kn+(k-1))\cdots(kn+1)}}{\frac{n^n}{1}}\right|$$

$$ = \frac{27}{e}\lim_{n\to\infty}\left|\frac{(n+1)^{k-1}(n+1)^{n+1}}{(kn+k)(kn+(k-1))\cdots(kn+1)n^n}\right|$$

From here, I consider the only the highest order term in the numerator and denominator, with there only being $k$ terms remaining from the factorial, yielding leading term $(kn)^k = k^kn^k$,

$$ = \frac{27}{e}\lim_{n\to\infty}\left|\frac{n^{k+n} + O(n^{k+n-1})}{k^kn^{k+n}+O(n^{k+n-1})}\right|$$

Solving the limit by dividing top and bottom by ${n^{k+n}}$ gives

$$ = \frac{27}{e}\frac{1}{k^k}$$

Which would be first less than 1 and therefore convergent at $k=3$, i.e. $\frac{27}{e}\frac{1}{3^3} = \frac{1}{e} < 1$. Then, by the ratio test, $k=3$ would be the first instance that this is true, because at $k=1$ we get $\frac{27}{e} > 1$ and at $k=2$ we get $\frac{27}{4e} > 1$, which diverge by the ratio test.

However, when I punch $\sum\limits_{n=1}^\infty\frac{(n!)^{3-1}(27n)^n}{(3n)!e^n}$ into Wolfram Alpha, it states that the series diverges, when the ratio test above shows that it converges for $k=3$. I'm not sure where my work is wrong and any pointers to the right answer of $k=4$ would be much appreciated. Thanks in advance!

Answer 1

We can write $$\begin{align} \frac{a_{n+1}}{a_n}&= \left(\frac{(n+1)!}{n!}\right)^{k-1}\frac{(27(n+1))^{n+1}}{(27n)^n}\frac{(kn+k)!}{(kn)!}\frac{e^n}{e^{n+1}}\\ &=(n+1)^{k-1}27(n+1)\left(1+\frac1n\right)^n\frac1e\frac1{(kn+k)\cdots(kn+1)}\\ &=27\left(1+\frac1n\right)^n\frac1e\frac{(n+1)^k}{(kn+k)\cdots(kn+1)}\\ &=27\left(1+\frac1n\right)^n\frac1e\frac1{k\left(k-\frac1n\right)\left(k-\frac2n\right)\cdots\left(k-\frac{k-1}{n}\right)}\\ &\to\frac{27}{k^k} \text{ as }n\to\infty \end{align}$$ Thus the series certainly converges when $k>3$ and diverges when $k<3$ and the case $k=3$ is doubtful.

For the case $k=3$, we have $$a_n=\frac{(n!)^2(27n)^n}{(2n)!e^n}\sim\frac{(27n)^n\sqrt{n\pi}}{4^ne^n}$$ by Stirling's approximation. Thus $a_n\to\infty$ as $n\to\infty$, and the series diverges for $k=3$.

Answer 2

$$a_n=\frac{(n!)^{k-1}\,(27n)^n}{(kn)!\,e^n}\implies$$ $$\log(a_n)=(k-1)\log(n!)+n \log(27n)-n-\log((kn)!)$$ Using Stirling approximation $$\log(a_n)=n (\log (27)-k \log (k))+\frac{1}{2} ((k-2) \log (2 \pi n)-\log (k))+\frac{(k-1) k-1}{12 k n}+O\left(\frac{1}{n^3}\right)$$ Apply it twice and continue with Taylor series to obtain $$\log(a_{n+1})-\log(a_n)=(\log (27)-k \log (k))+\frac{k-2}{2n}+O\left(\frac{1}{n^2}\right)$$ $$\frac{a_{n+1} } {a_n }=e^{\log(a_{n+1})-\log(a_n) }\sim 27 k^{-k}\, e^{\frac{k-2}{2n} }$$ So, you need $k>3$ for convergence.

If $k=3$, we have $$\frac{a_{n+1} } {a_n }=\frac{9 n^{-n} (n+1)^{n+2}}{e (9 n (n+1)+2)}\to 1$$ then divergence