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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.

UltraCommit
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  • My book says that there is a closed form. – UltraCommit Jul 13 '14 at 07:02
  • Please, give the answer from the book if it is given. Cheers :) – Claude Leibovici Jul 13 '14 at 07:09
  • The book says, "Calling f(x) the function sum of the given series, compute f''(3)". This means that f(x) exists in a closed form. – UltraCommit Jul 13 '14 at 07:11
  • @UltraCommit No, that doesn't mean that it necessarily has a closed form. I would bet that the book is encouraging you to think about the form of a power series. The terms tell you information about the derivatives. –  Jul 13 '14 at 07:17
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    $f(3)=\infty$ and, as T. Bongers wrote, the statement in the book does not mean that there is any closed form. In fact, $\infty$ is the result for any $x$ greater or equal to $e$. – Claude Leibovici Jul 13 '14 at 07:19
  • I've edited the question by adding the original text and my substitution y = 3-x. It's required to compute the second derivate in the point x = 3. – UltraCommit Jul 13 '14 at 07:21
  • You totally changed the problem ! – Claude Leibovici Jul 13 '14 at 07:23
  • Why I've totally changed the problem? Please explain... and solve, if you can! :-) – UltraCommit Jul 13 '14 at 07:24
  • With the (-1)^n at the beginning you can find the solution in a closed form?! – UltraCommit Jul 13 '14 at 07:25
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    Many consider it rude to vastly alter the question being asked, after people have already gone to the trouble of answering it. I strongly suggest that you revert the edit, and ask the new question separately (linking back to this question if you feel it's necessary). –  Jul 13 '14 at 07:40

1 Answers1

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.

Claude Leibovici
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