# How to calculate $\lim_{t \to \infty} \sum_{c=1}^{t-1} \frac{q^c}{t-c}$

How can I find the value of the infinite sum:

$$\lim_{t \rightarrow \infty} \sum_{c=1}^{t-1} \frac{q^c}{t-c}$$ where $$0 .

I tried putting the expression into WolframAlpha for some numerical experiments. It saw some behaviors that the infinite sum actually converges to 0. I am interested to know the reason behind this.

WolframAlpha also gave me this closed form expression for the summation involving Lerch Transcendent which I don't understand. Is there a simpler way to derive this result?

Any help is appreciated.

You want to analyse :

$$\sum_{k=1}^{t-1} \frac{q^k}{t-k}.$$

You can reverse the sum by reindexing $$k:=t-k$$ and it comes :

$$\sum_{k=1}^{t-1} \frac{q^{t-k}}{k}$$ that is equal to : $$q^t\sum_{k=1}^{t-1} \frac{(q^{-1})^k}{k}.$$

From the relation : $$-\log(1-x)=\sum_{k\geq1}t^k/k,$$

You have that in $$q^t\sum_{k=1}^{t-1} \frac{(q^{-1})^k}{k}.$$

the sum tends to : $$-\log(1-1/q)$$ that is finite.

Where as $$q^t$$ tends to 0.

Finally, the sum tends to 0.

• The relation with $\log(1-t)$ holds only when $|t|<1$, it does not hold with $q^{-1}$. I made the same mistake in my first reply... Mar 15, 2023 at 14:16

Let $$\mu$$ be the counting measure on $$\mathbb{N^*}$$, then the sum is $$$$S(t) = \int_{\mathbb{N^*}}\frac{q^c}{t-c}\mathbf{1}_{c\le t-1}d\mu(c)$$$$ but $$$$0\le \frac{q^c}{t-c}\mathbf{1}_{c\le t-1}\le q^c \in L^1(\mathbb{N^*}, d\mu)$$$$ and for every $$c\in \mathbb{N}^*$$ one has $$$$\lim_{t\to\infty}\frac{q^c}{t-c}\mathbf{1}_{c\le t-1} = 0$$$$ hence $$\displaystyle\lim_{t\to\infty}S(t) = 0$$ by the dominated convergence theorem.

Remark: Using the formula for $$n \in {\mathbb{N}}^{\ast }$$

$$$$\frac{1}{t-c} = \sum _{k = 1}^{n} \frac{{c}^{k-1}}{{t}^{k}}+\frac{1}{t-c} {\left(\frac{c}{t}\right)}^{n}$$$$

one gets

$$$$S \left(t\right) = \sum _{k = 1}^{n} \frac{1}{{t}^{k}} \left(\sum _{c = 1}^{t-1} {c}^{k-1} {q}^{c}\right)+\frac{1}{{t}^{n}} \sum _{c = 1}^{t-1} \frac{{c}^{n} {q}^{c}}{t-c}$$$$

We can now prove the following asymptotic expansion when $$t \rightarrow \infty$$

$$$$\boxed{S \left(t\right) = \sum _{k = 1}^{n} \frac{{a}_{k}}{{t}^{k}}+o \left(\frac{1}{{t}^{n}}\right)}$$$$

where

$$$${a}_{k} = \sum _{c = 1}^{\infty } {c}^{k-1} {q}^{c}=\text{Li}_{-k+1}(q)$$$$

Indeed we have

$$$$S \left(t\right)-\sum _{k = 1}^{n} \frac{{a}_{k}}{{t}^{k}} = -\sum _{k = 1}^{n} \frac{1}{{t}^{k}} \left(\sum _{c = t}^{\infty } {c}^{k-1} {q}^{c}\right)+\frac{1}{{t}^{n}} \sum _{c = 1}^{t-1} \frac{{c}^{n} {q}^{c}}{t-c}$$$$

The last sum tends to $$0$$ by Lebesgue's theorem, so the last term is indeed of order $$o \left({t}^{{-n}}\right)$$. Furthermore we have

$$$$\sum _{c = t}^{\infty } {c}^{k-1} {q}^{c} = {q}^{t} \sum _{j = 0}^{\infty } {\left(t+j\right)}^{k-1} {q}^{j} = {q}^{t} {P}_{k-1} \left(t\right) = o \left({t}^{{-n}}\right)$$$$

where $${P}_{k-1} \left(t\right)$$ is a polynomial of degree $$k-1$$, which concludes the proof.

In particular when $$n=1$$ we note that $$$$S \left(t\right) \sim \frac{{a}_{1}}{t} \sim {\frac{q}{1-q}}\times{\frac{1}{t}}$$$$

• $q^t \rightarrow 0$ but the sum diverges. It is unclear to me why the product goes to zero. Thanks Mar 15, 2023 at 13:23
• @antonio See the updated version, completely different. Mar 15, 2023 at 14:17
• The integral does not seem to go to zero. $\int_{0}^\infty q^x dx = - \frac{1}{\log q}$ Mar 15, 2023 at 14:36
• @antonio the function $q^c$ is only the dominating function that we need to apply Lebesgue's theorem. Its integral is constant, wich is always the case when this theorem is applied, but the function below the integral of $S(t)$ tends to $0$ pointwise when $t\to\infty$, hence this integral tends to $0$. Mar 15, 2023 at 14:39

It is more natural to consider $$S_n=\sum_{k=1}^n\frac{q^k}{n-k}=\int_0^1P_n(u,q)du$$ where $$P_n(u,q)=q^{n-1}+uq^{n-2}+u^2q^{n-3}+\cdots+u^{n-2}q+u^{n-1}=\frac{u^n-q^n}{u-q}.$$ Let us show $$S_n\to 0.$$ With $$p$$ such that $$q we write $$S_n=\int_0^q+\int_q^p+\int_p^1P_n(u,q)du=A_n+B_n+C_n.$$ We have $$A_n\leq q\times nq^{n-1},\ B_n\leq (p-q)\times np^{n-1}$$ and $$C_n= \int_p^1\frac{u^n-q^n}{u-q}du\leq\frac{1}{p-q}\int_p^1u^ndu=\frac{1}{p-q}\times \frac{1-p^{n+1}}{n+1}.$$ Clearly $$A_n,\ B_n,\ C_n$$ go to zero.

Actually $$(n+1)(A_n+B_n)\to 0$$ clearly. Since $$\frac{n+1}{1-q}\int_p^1(u^n-q^n)du\leq(n+1)C_n\leq\frac{n+1}{p-q}\int_p^1(u^n-q^n)du,$$ we have for all $$q $$\frac{1}{1-q}\leq \liminf(n+1)C_n\leq \limsup(n+1)C_n\leq \frac{1}{p-q}$$ which implies by doing $$p\to 1$$ that $$(n+1)C_n\to \frac{1}{1-q}.$$ As a result

$$\lim_{n\to \infty}nS_n=\frac{1}{1-q}.$$

• Thank you @Gérard! I get it now. However, may I ask a further question: What if the sum is now $S_n = \sum_{k=1}^n \frac{n q^k}{(n-k)^2}$? Thank you! Mar 20, 2023 at 11:44
• $nS_n\to q/(1-q)$ by dominated convergence. Indeed $f_n(k)=\frac{n^2}{(n-k)^2} q^k$ if $k=0.\ldots,n-1$ and $f_n(k)=0$ is dominated by $g(k)=(k+1)^2q^k.$This quick proof is an adaptation of an idea of my colleague jandri. Mar 21, 2023 at 16:35