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Does there exist a closed form for finite summation of the sequence $\sum^n_{i=1}{e^i/i}$ ?

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Yes it does. Is that a homework problem? – AD. Feb 13 '13 at 20:04
I think it is fair to say that there is no reasonably elementary closed form for this value. – copper.hat Feb 13 '13 at 20:26
Considering that even harmonic numbers have no known simple elementary closed form despite of hundreds of years of intense inspection, I can hardly expect that it would do so. – Sangchul Lee Nov 26 '13 at 22:46

Take the derivative of the function $f(x) = \sum_{i = 1}^n \frac{x^i}{i}$ which is easier to compute.

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Uh? No matter what $\,n\,$ is chosen?! – DonAntonio Feb 13 '13 at 20:04
@DonAntonio I think $e=0$, right. – AD. Feb 13 '13 at 20:06
You are right, I edit. – Damien L Feb 13 '13 at 20:06
And how do you integrate the result? – copper.hat Feb 13 '13 at 20:09
I still can't see how that sum is "easier" to compute...perhaps an integral, though...? – DonAntonio Feb 13 '13 at 20:09


Integrate indefinitely the above:

$$\int f(x)\,dx=\sum_{k=1}^n\int x^{k-1}dx=\sum_{k=1}^n\frac{x^k}{k}\ldots$$

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How do you extract the result from the integral? – copper.hat Feb 13 '13 at 20:14
Integrate the RHS of the first line wrt $\,x\,$ ... – DonAntonio Feb 13 '13 at 20:16
An formula involving an integral isn't really much more of a closed form than a formula involving a summation. – Hurkyl Feb 13 '13 at 20:17
Well @Hurkyl, that may depend on the OP, but I'd agree with you in general. Anyway that's an option for him... – DonAntonio Feb 13 '13 at 20:19
@DonAntonio: I understand the underlying idea, but what integration limits do you use? (Or are you integrating in $\mathbb{C}$?) – copper.hat Feb 13 '13 at 20:20

Maple writes it as $$\sum_{i=1}^n \frac{{\rm e}^i}{i} = -\Phi \left( {{\rm e}},1,n \right)\; {{\rm e}^{n}}-\ln \left( -{{\rm e}}+1 \right) +{\frac {{{\rm e}^{n}}}{n}}$$

where $\Phi$ is the Lerch Phi function.

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Lerch at $(z,s,n)$ with $s=1$ and $|z|\gt1$... Are you sure that this simply exists? Probably related: the argument of the logarithm on the RHS is real negative? – Did Oct 26 '13 at 8:38
Defined using analytic continuation. – Robert Israel Oct 27 '13 at 2:09
You mean, the logarithm on the RHS? But surely you realize that the LHS is purely real? – Did Oct 27 '13 at 6:59
Yes, I do realize that. So what? A real number can be written as the sum of non-real complex numbers. – Robert Israel Oct 27 '13 at 8:05
Sure but then one should explain the determination of the logarithm one is using. Another problem is to define $\Phi(z,s,a)$ when $|z|\gt1$, and this probably involves another choice of cut set. As long as you say nothing about these two points, I am afraid this answer is not very useful. – Did Oct 27 '13 at 8:44

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