Upper bound on $\sum_{k=1}^T \frac{1}{k (1+a)^{T-k}}$

Is there any reasonable upper bound for the following quantity $$\sum_{k=1}^T \frac{1}{k (1+a)^{T-k}}$$

where $$a>0$$ with respect to $$T$$ and $$a$$ (something like $$\mathcal{O}(\frac{\log (T)}{aT}$$)? I tried to compute integral $$\int_{0}^T \frac{1}{x (1+a)^{T-x}}dx,$$ which should be upper bound on this sum as $$f(x) = \frac{1}{x (1+a)^{T-x}}$$ is decreasing on $$(0, T)$$, but I did not achieve to get reasonable expression.

2 Answers

Changing variables to $$n=T-k$$, and letting $$x=1+a$$, we can write the sum as $$S=\sum_{n=0}^{T-1}\frac{x^{-n}}{T-n}=\frac{1}{T}+\frac{x^{-1}}{T-1}+\frac{x^{-2}}{T-2}+\cdots+\frac{x^{1-T}}{1}.$$ Recognizing that the right side of $$Sx^T=x+\frac{x^2}{2}+\cdots +\frac{x^T}{T}$$ would be the $$T^{th}$$ Taylor polynomial of $$-\log(1-x)$$, except for the annoying fact that $$x>1$$ so the Taylor series does not converge here.

Instead, the series behavior geometrically and is dominated by its last few terms. One way to obtain a somewhat reasonable and explicit upperbound is by keeping the last term and decreasing the denominators of all other terms to $$1$$, then summing the geometric series to obtain $$Sx^T\leq x+x^2+\cdots+x^{T-1}+\frac{x^T}{T}=\frac{x^T-1}{x-1}-1+\frac{x^T}{T},$$ showing that $$S\leq \frac{1}{a}(1-(1+a)^{-T})-1+\frac{1}{T}.$$ To gauge how tight of an upper bound this is would require knowing how $$a$$ compares to $$1$$, and different procedures could be used in the regimes when $$a$$ is very close to $$0$$ vs when $$a$$ is much larger than $$1$$.

Here is a fairly loose upper bound. (I wrote this for the cross validated posting and am copying it here since that one was deleted.)

\begin{aligned} \sum_{k=1}^T\frac{1}{k(1+a)^{T-k}} &= (1+a)^{-T} \sum_{k=1}^T\frac{(1+a)^k}{k} \\ &\le (1+a)^{-T} \sum_{k=1}^T(1+a)^k \\ &= (1+a)^{-T} \frac{(1-(1+a)^{T+1})}{1-(1+a)}\\ &= \frac{1}{a}(1+a)^{-T} \left((1+a)^{T+1}-1\right)\\ &= \frac{1}{a}\left((1+a)-(1+a)^{-T}\right)\\ \end{aligned}