$$\frac{1}{1-x} = 1+x+x^2+x^3+ \cdots$$

Is there a $f(x)$ that has the series of $n$th roots?

$$f(x)= x+x^{\frac{1}{2}}+x^{\frac{1}{3}}+ \cdots$$

Wolfram Alpha seemed to not understand my input.


Your series doesn't converge because as Hagen von Eitzen said $\sqrt[n]x\to 1$ as $n\to\infty$.

Though if you consider:


As $n\to\infty$ this does indeed converge and has the alternate series expansion:


Where $\zeta$ is the Riemann zeta function, $\zeta(s)=\sum_{k=1}^\infty\frac{1}{k^s}$.

Similar looking series also crop up in expansions for prime counting functions. For example the Von Mangoldt function is usually used as a proxy for counting primes due to it's "nice" arithmetic nature and once an expression has been obtained for a sum involving the Mangoldt function. Inverting it back into a series involving only primes requires introducing many integer roots.

For instance note the two very good asymptotic approximations for the prime counting function $\pi$ and the first Chebyshev function $\vartheta$ given by these similar looking series:

$$\sum_{p\leq x}1=\pi(x)\sim \sum_{m=1}^\infty\frac{\ln(x)^m}{mm!\zeta(m+1)}$$ $$\sum_{p\leq x}\ln(p)=\vartheta(x)\sim \sum_{m=2}^\infty\frac{\ln(x)^m}{m!\zeta(m)}$$

Which appear as a result of their inversions with other functions related to the Mangoldt function:



Where we have that:

$$\Pi(x)=\sum_{2\leq n\leq x}\frac{\Lambda(n)}{\ln(n)}$$ $$\psi(x)=\sum_{n\leq x}\Lambda(n)$$

With $\Lambda$ the Von Mangoldt function appearing in the above partial sums.

Though I assume by your MSE picture and name that your probably already familiar with a lot of this stuff, as the error terms in the series' can be written as sums over zeros of the zeta function.

  • $\begingroup$ @anon Yea my bad didn't notice that, I changed it so it works now. $\endgroup$ – Ethan Feb 6 '14 at 2:28
  • $\begingroup$ @Ethan Yes, I am somewhat familiar with the zeta function; no formal education on it...yet. Thanks for your informative answer! $\endgroup$ – zerosofthezeta Feb 6 '14 at 4:57

For $x>0$ we have $\sqrt[n]x\to 1$ hence the series cannot converge

  • 3
    $\begingroup$ What's more, you can show with some effort that for large $n$, $\sqrt[n]{x}\approx 1+\frac Cn$ for some $C$, so trying to sum $\left(\sqrt[n]{x}-1\right)$ doesn't turn out any better. $\endgroup$ – Steven Stadnicki Feb 5 '14 at 21:08
  • $\begingroup$ @RahulNarain It has to, by the usual alternating series convergence theorems. Or do you mean that it converges to a 'known' function of $x$? $\endgroup$ – Steven Stadnicki Feb 5 '14 at 21:27
  • $\begingroup$ @Steven: Not really sure what my point was. I had forgotten the usual theorems. It's been a while since I did real analysis... $\endgroup$ – Rahul Feb 5 '14 at 21:30

This is not exactly an answer... but I find the question intriguing and totally want to know if it's been studied before. Who cares if it doesn't converge?

What is interesting about your series is that because the denominators of the exponents are not bounded, it is not a Puiseux series, nor even a Hahn series (which I had never heard of) because the set of exponents is not well-ordered!


For $x>0,~x+x^{\frac{1}{2}}+x^{\frac{1}{3}}+...$ can't converges since $x^{\frac{1}{n}}\to1.$ Neither does it for $x<0$ since $x^{\frac{1}{2}}$ is undefined then. So the only possibility is the trivial one: $$f:\{0\}\to\mathbb R:0\mapsto 0$$


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.