0
$\begingroup$

Is it possible to simplify this series?

$$ f(x) = \sum\limits_{k=1}^n \frac{x^k}{k} $$

If I derive it, I get the geometric series

$$ f'(x) = \sum\limits_{k=1}^n x^{k-1} = \sum\limits_{k=0}^{n-1} x^{k} = \frac{1-x^n}{1-x} $$

But this seems to lead to nothing as

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

can't expressed simpler as a comment in my previous question Antiderivative of $\frac{1-x^n}{1-x}$ mentioned.

I know, that $\lim\limits_{n \to \infty} \int \frac{1-x^n}{1-x} dx = -\ln{(|x-1|)}, \quad -1 < x < 1$

But what about without these restrictions of $n$ and $x$?

$\endgroup$
2
  • $\begingroup$ Well, I did already (second equation). $\endgroup$ Commented Dec 3, 2020 at 19:24
  • $\begingroup$ There's no hope for a really simple closed form for this function, because such a closed form would imply a really simple closed form for $f(1) = \sum_{k=1}^n \frac{1}{k} = H_n$, the harmonic number (en.wikipedia.org/wiki/Harmonic_number). $\endgroup$ Commented Dec 3, 2020 at 21:16

2 Answers 2

1
$\begingroup$

Beside the incomplete Beta function, you also have $$f(x) = \sum\limits_{k=1}^n \frac{x^k}{k}=-x^{n+1}\, \Phi (x,1,n+1)-\log (1-x)$$ where appears the Lerch transcendent function.

$\endgroup$
0
$\begingroup$

Mathematica tells me it is $$ f(x) = \sum_{k=1}^n \frac{x^k}{k} = \int\frac{1-x^n}{1-x}\ \mathrm{d}x = -\mathrm{B}(x,1+n,0) - \ln(1 - x),$$ where $\mathrm{B}(x,a,b)$ is the incomplete Beta function. Not sure if that satisfies you. Also not sure what you mean by simplifying.

$\endgroup$
2
  • $\begingroup$ Thanks for your answer. But sadly thats not what I expected. The incomplete beta functions doesn't seem to be "simple". I mean expressions like $\frac{1}{1-x}$ or something with logarithm. Because all the other derivatives of the geometric series can be simplified easy, but the antiderivative, too? $\endgroup$ Commented Dec 3, 2020 at 19:48
  • $\begingroup$ No. If there's no closed form there's no closed form. Basically the function is defined by the integral. $\endgroup$ Commented Dec 3, 2020 at 19:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .