# Computing the sum of an infinite series as a variant of a geometric series

I came across the following series when computing the covariance of a transform of a bivariate Gaussian random vector via Hermite polynomials and Mehler's expansion:

$$S = \sum_{n=1}^{\infty} \frac{\rho^n}{n^{1/6}}$$ for $$\vert \rho \vert < 1$$. We know that $$S$$ must be finite and satisfy $$S \le \rho (1-\rho)^{-1}$$ since the original series is dominated by $$\sum_{n=1}^{\infty} \rho^n$$.

However, there is a catch if we use for $$S$$ the upper bound $$\rho (1-\rho)^{-1}$$, which tends to $$\infty$$ when $$\rho \to 1-$$. This happens when the two marginal random variables in the Gaussian vector are almost surely, positively linearly dependent (asymptotically).

So, the target is to obtain a good upper bound, much better than $$\rho (1-\rho)^{-1}$$ when we restrict $$\rho$$ to be away from $$1$$, to reduce the effect of $$\rho \to 1-$$. In other words, let $$1-\rho = \delta$$ for some fixed $$\delta \in (0,1)$$, what is a better upper bound for $$S$$?

Because of the scaling term $$n^{-1/6}$$ that induces a divergent series $$\sum_{n=1}^{\infty} n^{-1/6}$$, probably not much improvement should be expected. I have Googled but did not find an illuminating technique for this. Any pointer or help is appreciated. Thank you.

Let $$r:=\rho$$, $$S(r):=S$$, and $$a:=1/6$$. Let us show that $$\begin{equation*} S(r)< Cr(1-r)^{a-1}\quad\text{and}\quad S(r)\sim C(1-r)^{a-1},\quad\text{where}\quad C:=\Gamma(1-a). \tag{0} \end{equation*}$$ Everywhere here, the equalities and inequalities are for $$r\in(0,1)$$, and the limit relations are for $$r\uparrow1$$.

Indeed, we have $$\begin{equation*} \frac1{n^a}=\frac1{\Gamma(a)}\,\int_0^\infty u^{a-1}e^{-nu}\,du \end{equation*}$$ and hence $$\begin{equation*} S(r)=\frac1{\Gamma(a)}\,\int_0^\infty du\, u^{a-1}\,\sum_{n=1}^\infty r^n e^{-nu} =\frac r{\Gamma(a)}\,I(r), \end{equation*}$$ where \begin{align*} I(r)&:=\int_0^\infty du\, \frac{u^{a-1} e^{-u}}{1-r e^{-u}} \\ & =\int_0^\infty du\, \frac{u^{a-1}}{e^u-r} \\ & <\int_0^\infty du\, \frac{u^{a-1}}{1+u-r} \\ & =(1-r)^{a-1}\int_0^\infty dv\, \frac{v^{a-1}}{1+v} \\ &=\Gamma(1-a)\Gamma(a)(1-r)^{a-1}, \end{align*} where we used the substitutions $$u=(1-r)v$$ and then $$\frac1{1+v}=t$$. So, the inequality in (0) is proved.

Also, \begin{align*} (1-r)^{1-a}I(r)& =\int_0^\infty dv\, \frac{v^{a-1}}{1+(e^{(1-r)v}-1)/(1-r)} \\ &\to\int_0^\infty dv\, \frac{v^{a-1}}{1+v} , \end{align*} by dominated convergence. So, the asymptotic relation in (0) is proved.

As seen from the above proof, the results in (0) hold for any $$a\in(0,1)$$.

• thank you for exposing the mechanisms of estimating $S$. Following your strategy, i did my computation by splitting the limits of the integral into 3 parts, found out that only the integral on the range $u = o(1)$ contributes to the inflation for the magnitude of $S$, and realized that $C (1 - \rho_0)^{-5/6}$ for a constant $C>0$ is the best one can get even when $\rho \le \rho_0 <1$. – Chee Jun 25 at 0:17
• I have now also given an asymptotically exact upper bound. – Iosif Pinelis Jun 25 at 10:35

Your function is the Polylogarithm function $$Li_{1/6}(\rho)$$. Mathematica indicates the correct asymptotic (at least for real $$\rho$$) is $$\Gamma(5/6)/(1-\rho)^{5/6}+O(1)$$.

• when you said the "correct asymptotic", did you mean the asymptotic when $\rho \to 1-$? Let us restrict $\rho$ to be away from $1$, what is a better upper bound on $S$? Thank you. – Chee Jun 24 at 22:45
• Anyway, your pointer is enough for me to work more on it. Thank you. – Chee Jun 24 at 22:53