Please help me to compute the sum:
$$\sum_{n=1}^{\infty } \frac{n!}{n^{n}} x^{n}$$
in a closed form. === here ends the original post.
After a few minutes I've added the following information:
This was the original problem:
$$\sum_{n=1}^{\infty } (-1)^{n}. \frac{n!}{n^{n}}. (x-3)^{n}$$
With the substitution y = (3-x), it becomes the text I've proposed.
It is required to compute explicitally f(x), and then to determinate the value of f''(3), this means the values of the SECOND DERIVATIVE in the point x = 3.
The original series is centered on x=3.
I would really like to know the closed form given in the book.
The following answer was given for the original post.
The only thing I have been able to do to arrive to something is to replace $n!$ by Stirling approximation, that is to say $$n! \simeq \sqrt{2 \pi } e^{-n} n^{n+\frac{1}{2}}$$ and using this, the result of the summation is given by $$\sum_{n=1}^{\infty } \frac{n!}{n^{n}} x^{n}\simeq \sqrt{2 \pi } \sum_{n=1}^{\infty } \sqrt{n} \Big(\frac{x}{e}\Big)^n=\sqrt{2 \pi } \text{Li}_{-\frac{1}{2}}\left(\frac{x}{e}\right)$$ which seems to be a quite reasonable approximation.
For the new problem $$f(x) = \sum_{n=1}^{\infty } (-1)^{n} \frac{n!}{n^{n}} (x-3)^{n}$$ $$f'(x) = \sum_{n=1}^{\infty }(-1)^n n^{1-n} n! (x-3)^{n-1}$$ $$f''(x) = \sum_{n=1}^{\infty }(-1)^n (n-1) n^{1-n} n! (x-3)^{n-2}$$
All terms are equal to $0$ except for $n=2$. Then ...
Added later
You even do not need to compute all of the above. Just write $$f(x) = \sum_{n=1}^{\infty } a_n (x-3)^{n}$$ Derive twice wrt $x$; you are just left with $2a_2$ and now replace $a_2$ by its definition.