Asymptotics of $\Sigma x^n/n^p$

Question

I am looking for asymptotics of

$$\sum_{n=1}^N \dfrac{x^n}{n^p}$$

in terms of $N$, where $x>1$ and $p$ is a positive number.

So I am looking for a $O(f(N))$ type result. For example what is the approximate value of $\sum_{n=1}^N \dfrac{5^n}{n^2}$ for large $N$. One possibility is probably something like $\frac{5^N}{N^2} (a+\frac{b}{N}+\frac{c}{N^2}+\cdots)$.

Answer 1

By Taylor expansion, we see that

$$\sum_{n=1}^N\frac{x^n}{n^p}=\sum_{k=1}^\infty\frac1{k!}\sum_{n=1}^N\frac{(n\ln x)^k}{n^p}=\sum_{k=1}^\infty\frac{(\ln x)^k}{k!}\sum_{n=1}^Nn^{k-p}$$

We then have by the Euler-Maclaurin formula:

$$\sum_{n=1}^Nn^{k-p}=\frac{N^{k-p+1}}{k-p+1}+\frac{N^{k-p}}2+\zeta(k-p)+\frac{(k-p)N^{k-p-1}}{12}+\mathcal O(N^{k-p-3})$$

where $\zeta$ is the Riemann zeta function. Thus, we end up with

$$\sum_{k=1}^\infty\frac{(\ln x)^k}{k!}\left(\frac{N^{k-p+1}}{k-p+1}+\frac{N^{k-p}}2+\zeta(k-p)+\frac{(k-p)N^{k-p-1}}{12}+\mathcal O(N^{k-p-3})\right)$$

---

By the Abel summation formula with $\phi(n)=\frac1{n^p}$ and $a_n=x^n$, we get

$$\sum_{n=1}^N\frac{x^n}{n^p}=\frac{1-x^{N+1}}{(1-x)N^p}+p\int_1^N\frac{1-x^{\lfloor t\rfloor+1}}{(1-x)t^{p+1}}\ dt\\=\frac1{1-x}\left(\frac{2-x^{N+1}}{N^p}-1-\int_1^N\frac{x^{\lfloor t\rfloor+1}}{t^{p+1}}\ dt\right)$$

And the last integral may be approximated.

Answer 2

We find \begin{align*} \sum_{n=1}^N \frac{x^n}{n^p} &= \sum_{n=1}^\infty \frac{x^n}{n^p} - \sum_{n=N+1}^\infty \frac{x^n}{n^p} \\ &= \sum_{n=1}^\infty \frac{x^n}{n^p} - x^{N+1}\sum_{k=0}^\infty \frac{x^k}{(k+N+1)^p} \qquad (\textrm{let } n=N+1+k) \\ &= \mathrm{Li}_p(x) - x^{N+1}\Phi(x,p,N+1) \qquad \textrm{(polylogarithm and Lerch transcendent)} \\ &\sim \frac{x^{N+1}}{x-1} \frac{1}{(N+1)^p}. \end{align*} In the last step we use the known asymptotics for the Lerch transcendent. Higher order corrections can be found. See, for example, Chelo Ferreira, José L. López, Asymptotic expansions of the Hurwitz–Lerch zeta function, Journal of Mathematical Analysis and Applications, Volume 298, Issue 1, 2004, Pages 210-224, ISSN 0022-247X, http://dx.doi.org/10.1016/j.jmaa.2004.05.040.

Addendum

As @martycohen notes, the expansion in the paper is for $x\in\mathbb{C}-[1,\infty)$. It should be possible to make a complex analytic argument that the asymptotic expansion above is valid for $x>1$. Rather than do that, instead I will present a more elementary approach.

Let $1<k<N$. Then $$\sum_{n=1}^N\frac{x^n}{n^p} = \sum_{n=N-k}^N\frac{x^n}{n^p} + \sum_{n=1}^{N-k-1}\frac{x^n}{n^p}.$$ But $$\sum_{n=1}^{N-k-1}\frac{x^n}{n^p} < \sum_{n=1}^{N-k-1}x^n = \frac{x^{N-k}-x}{x-1}$$ and $$\sum_{n=1}^{N-k-1}\frac{x^n}{n^p} > \frac{1}{N^p}\sum_{n=1}^{N-k-1}x^n = \frac{1}{N^p}\frac{x^{N-k}-x}{x-1}$$ and so $$\sum_{n=1}^{N-k-1}\frac{x^n}{n^p} = O(x^{N-k}).$$ Therefore, $$\sum_{n=1}^N\frac{x^n}{n^p} = \sum_{n=N-k+1}^N\frac{x^n}{n^p} + O(x^{N-k}).$$

Answer 3

Just playing around to see what happens.

If $x > 1$ then

$\begin{array}\\ \sum_{n=1}^N \dfrac{x^n}{n^p} &=x^N\sum_{n=1}^N \dfrac{x^{n-N}}{n^p}\\ &=x^N\sum_{n=0}^{N-1} \dfrac{x^{-n}}{(N-n)^p}\\ &=\dfrac{x^N}{N^p}\sum_{n=0}^{N-1} \dfrac{x^{-n}}{(1-n/N)^p}\\ &=\dfrac{x^N}{N^p}\sum_{n=0}^{N-1} x^{-n}(1-n/N)^{-p}\\ &=\dfrac{x^N}{N^p}\left(1+\dfrac1{x(1-1/N)^p}+\dfrac1{x^2(1-2/N)^p}+...\right)\\ \end{array} $

To see how (or if) this converges, look at the ratio of consecutive terms.

$\begin{array}\\ \dfrac{x^{-n-1}(1-(n+1)/N)^{-p}}{x^{-n}(1-n/N)^{-p}} &=\dfrac{(N-n)^p}{x(N-n-1)^{p}}\\ &=\dfrac{1}{x(1-1/(N-n))^{p}}\\ &\approx\dfrac{1}{x(1-p/(N-n))} \qquad\text{for small } n\\ &\approx\dfrac{1}{x}(1+p/(N-n))\\ &\approx\dfrac{1}{x}(1+p/(N(1-n/N))\\ &\approx\dfrac{1}{x}(1+(p/N)(1+n/N))\\ &=\dfrac{1}{x}(1+p/N+pn/N^2)\\ &\approx\dfrac{1}{x}(1+p/N)\\ \end{array} $

This is $1$ when $x(1-1/(N-n))^{p} =1$ or $1-1/(N-n) =x^{-1/p} $ or $N-n =\dfrac{1}{1-x^{-1/p}} $.

Using the approximate ratio of $\approx\dfrac{1+p/N}{x} $, the approximate expansion is $\dfrac{x^N}{N^p}\left(1+\dfrac{1+p/N}{x}+\dfrac{(1+p/N)^2}{x^2}+...\right) $.

That's as far as I can take it now.

Hope this helps.