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I am testing the convergence the following series $$\sum_{n=1}^\infty\frac{n^n}{(n!)^2},$$ using intuitive methods (without using ratio test). For instance, suppose that $n$ is even. If I write down: $$\frac{n^n}{(n!)^2}=\frac{1}{n}\cdot\frac{n}{(n-1)^2}\cdot\frac{n}{(n-2)^2}\cdots\frac{n}{(n-n/2)^2}\cdots\frac{n}{3}\cdot\frac{n}{2}\cdot n$$ and I take the pairwise product first $\times$ last, second $\times$ second to last etc..., I get that $$\frac{n^n}{(n!)^2}=\prod_{k=1}^{n/2}\frac{n}{k}\frac{n}{(n-k+1)^2}.$$ Now for large enough $n$, each term in the product is close to $1/k$. Therefore my intuition is that $$\frac{n^n}{(n!)^2}\sim \prod_{k=1}^{n/2}\frac{1}{k}=\frac{1}{(n/2)!}.$$

Is the intuition true? And if yes can it be turned onto a proof?

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  • $\begingroup$ Or: $\frac {n^n}{n!}$ is one of the positive summands in $e^n=\sum_k\frac{n^k}{k!}$, hence $\frac{n^n}{n!}<e^n$. On the other hand, $n!$ is the product of $(n-2)$ factors $\ge 3$, so $n!\ge \frac29 3^n$, so $\frac{n^n}{n!^2}<\frac 92\cdot(e/3)^n$. $\endgroup$ – Hagen von Eitzen Sep 15 at 13:50
  • $\begingroup$ Stirling's approximation, or the proof of it, might be helpful here. $\endgroup$ – Matthew Daly Sep 15 at 13:52
  • $\begingroup$ Thank you for the comments! I am already aware of both approaches. My question is whether my intuition is true or not, and whether it can be continued to a proof. $\endgroup$ – UserA Sep 15 at 14:01
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This is not an answer but the result of some lucubration of mine.

Simply Beautiful Art gave a good answer to your question.

Nevertheless, I considered the behavior of $$f(n)=\frac{n^n }{(n!)^2}\left(\left(\frac{n}{p}\right)!\right)^q$$ Using Stirling approximation and Taylor series for large values of $n$, the leading term disappears if $p=q=e$.

Using these values, it seems that $f(n)\sim a n^b$ with $a \sim \frac 12$ and $b \sim \frac{117}{326}$. So, a useless approximation could be $$\frac{n^n }{(n!)^2}\sim \frac{2 n^{-\frac{117}{326}}} {\left(\left(\frac{n}{e}\right)!\right)^e}$$ The table below gives, for a few values of $n$, the logarithms of lsh and rhs. $$\left( \begin{array}{ccc} 10 & -7.18297 & -7.22237 \\ 20 & -24.7566 & -24.7738 \\ 30 & -47.2806 & -47.2904 \\ 40 & -73.0861 & -73.0923 \\ 50 & -101.354 & -101.358 \\ 60 & -131.596 & -131.598 \\ 70 & -163.483 & -163.485 \\ 80 & -196.784 & -196.785 \\ 90 & -231.322 & -231.323 \\ 100 & -266.962 & -266.961 \\ 110 & -303.593 & -303.592 \\ 120 & -341.126 & -341.125 \\ 130 & -379.486 & -379.485 \\ 140 & -418.611 & -418.609 \\ 150 & -458.445 & -458.443 \\ 160 & -498.942 & -498.940 \\ 170 & -540.060 & -540.058 \\ 180 & -581.764 & -581.762 \\ 190 & -624.020 & -624.018 \\ 200 & -666.801 & -666.798 \end{array} \right)$$

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Since the amount of terms is not fixed, it does not suffice to simply observer the nature of the terms in the given product. The product also seems to be incorrect. Note that your expansion is missing some squares in the denominator and when you write it out in product notation, you are still missing a square in the denominator.

While you say you don't want to use the ratio test, looking at the ratio is not only easier, but allows you to develop some simple asymptotic expansions. For example, note that

$$\frac{(n+1)^{n+1}}{(n+1)!}\times\frac{n!}{n^n}=\left(1+\frac1n\right)^n\underset{n\to\infty}\longrightarrow e$$

which gives the weak Stirling approximation:

$$\frac{n^n}{n!}\approx e^n$$

which is sufficient to tackle your problem. To make this more rigorous, it would suffice to use something such as

$$2\le\left(1+\frac1n\right)^n\le3$$

and hence

$$2^{n-1}\le\frac{n^n}{n!}\le3^{n-1}$$

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From the Lagrange inversion theorem we have

$$ \sum_{n\geq 1}\frac{n^n}{n!}x^n = \frac{-W(-x)}{1+W(-x)} $$ (see also this answer) for any $|x|<\frac{1}{e}$, with $W(x)$ being Lambert's function, i.e. the inverse function of $x e^x$ in a neighbourhood of the origin. For any $n\geq 7$ we have $n!\geq 3^n $, hence $$ \sum_{n\geq 7}\frac{n^n}{n!}\cdot\frac{1}{n!}\leq \sum_{n\geq 7}\frac{n^n}{n!}\left(\frac{1}{3}\right)^n $$ and your series is clearly convergent.

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