7
$\begingroup$

I am trying to evaluate this infinite series that looks beautiful to me $$S=\sum_{n=1}^{\infty} \frac{x^n}{1+x+...+x^n}$$ All I could do is prove that it converges for $|x|<1$ and I tried to find a suitable form so that I can exploit it with an integral, however I was not successful. I would love to see a closed form. (this is a question that was also posted on other sites like AoPS but it didnt receive an answer so I hope no one minds if I post here too)

$\endgroup$
  • 5
    $\begingroup$ This is $$\sum_{n=1}^\infty(d(n)-d(n-1))x^n$$ where $d(0)=0$ and, for every $n\geqslant1$, $d(n)$ is the number of divisors of $n$, also denoted $\sigma_0(n)$ or $\sigma(n)$. $\endgroup$ – Did Apr 20 '18 at 17:09
  • $\begingroup$ @Did Interesting, so this is the generating function for $d(n) - d(n-1)$. Do you happen to have any combinatorial insight why this would be true? $\endgroup$ – user159517 Apr 20 '18 at 17:11
  • $\begingroup$ Algebraically, this follows readily from the identity $$1+x+\cdots+x^n=\frac{1-x^{n+1}}{1-x}$$ Combinatorially, one would first have to explain the series in the question combinatorially... $\endgroup$ – Did Apr 20 '18 at 17:13
  • 1
    $\begingroup$ Please look at the OEIS sequence A051950 generating function. $\endgroup$ – Somos Apr 20 '18 at 18:08
  • $\begingroup$ I get $\sum_{n=1}^\infty \left(d(n+1)-d(n)\right)x^n$ $\endgroup$ – ccorn Apr 20 '18 at 18:13
6
$\begingroup$

We can make use of the Lambert series:

$$S(x)=\sum_{n=1}^{\infty}\frac{x^n}{\sum_{k=0}^nx^k}=\sum_{n=2}^{\infty}\frac{x^{n-1}-x^n}{1-x^n}=\left(\frac{1}{x}-1\right)L(1,x)-1,$$

where $L(f,q)$ is the Lambert generating function of $f$:

$$L(f,q):=\sum_{n=1}^{\infty}\frac{f(n)q^n}{1-q^n}=\sum_{n=1}^{\infty}(f*1)(n)q^n$$

and $*$ is Dirichlet convolution. Plugging in $L(1,q)=\frac{\psi_q(1)+\ln(1-q)}{\ln(q)}$, where $\psi_q(z)$ is the $q$-digamma function, we have

$$S(x)=\left(\frac{1}{x}-1\right)\frac{\psi_x(1)+\ln(1-x)}{\ln(x)}-1.$$

If you want a power series you may write it, as @did pointed out, as

$$S(x)=\sum_{n=1}^{\infty}(\sigma_0(n+1)-\sigma_0(n))x^n,$$

where $\sigma_0(n)$ is the divisor function which counts the number of divisors of $n$.

$\endgroup$
  • $\begingroup$ i must say :D the answers are totally marvellous :D simply and straight $\endgroup$ – Jose Garcia Apr 20 '18 at 18:09
  • $\begingroup$ @ccorn Oops that’s a typo. Will fix it. $\endgroup$ – Jacob Apr 20 '18 at 18:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.