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

Justin
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3 Answers3

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

Damien L
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$$f(x):=\sum_{k=1}^nx^{k-1}=\sum_{k=0}^{n-1}x^k=\frac{1-x^{n-1}}{1-x}$$

Integrate indefinitely the above:

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

DonAntonio
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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
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    An formula involving an integral isn't really much more of a closed form than a formula involving a summation. –  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
  • None @copper.hat: indefinite integral, unless the OP adds some more conditions, of course. – DonAntonio Feb 13 '13 at 20:20
  • Then I don't understand, presumably $e = \exp(1)$? – copper.hat Feb 13 '13 at 20:21
  • @copper: Indefintie integrals don't have limits; the method described just says to antidifferentiate. You use a known value ($x=0$) to fix the constant afterwards. The plug in $x=e$ if you want the value of the sum at $x=e$. –  Feb 13 '13 at 20:21
  • @Hurkyl: Like I say, I understand the basic idea. The details seem a little more complicated... – copper.hat Feb 13 '13 at 20:22
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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.

Robert Israel
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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 08:38
  • Defined using analytic continuation. – Robert Israel Oct 27 '13 at 02:09
  • You mean, the logarithm on the RHS? But surely you realize that the LHS is purely real? – Did Oct 27 '13 at 06:59
  • 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 08:05
  • 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 08:44
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    Once you choose a branch of log, you use the corresponding branch of $\Phi$ that makes $\Phi(z,1,n) z^n + \ln(1-z)$ real when $z$ is a positive real. So if you use the principal branch of log, the branch cut for $\Phi$ is on the real axis from $1$ to $+\infty$. This does seem to be the choice Maple makes. – Robert Israel Oct 27 '13 at 16:36