Suppose we are given a fraction $\alpha <1$. My question is whether we can derive an upper bound on summations of the following form:
$$S_t= \sum_{i=1}^{t} \frac{\alpha^{t-i}}{\sqrt{i}},$$ where $\alpha <1$.
I have tried to tackle this by bounding the summation by an integral. But it then boils down to bounding integrals of the form $ \int_{x=0}^{t} e^{x^2} dx $, for which I am not sure if there are known bounds (other than using complimentary/imaginary error functions). Summation by parts also doesn't seem to help much. Any pointers on how we can go about bounding $S_t$ would be very helpful. My conjecture is that $S_t = O(t^{-\beta})$ for some suitable $\beta >0$ (most likely $\beta=1/2$) based on Matlab plots.
Start from the identity $$ \frac 1{\sqrt j} = \frac 1{\sqrt \pi}\int_{-\infty}^{+\infty}e^{-j x^2}dx $$ to obtain $$ \sum_{j=1}^t\frac {a^{t-j}}{\sqrt j} = \frac 1{\sqrt \pi}\int_{-\infty}^{+\infty}\frac{a^t-e^{-t x^2}}{a e^{x^2}-1}dx. $$ Change variable $y^2=tx^2$ to obtain $$ \sum_{j=1}^t\frac {a^{t-j}}{\sqrt j} = \frac 1{\sqrt {\pi t}}\int_{-\infty}^{+\infty}\frac{a^t-e^{-y^2}}{a e^{y^2/t}-1}dy. $$ It should be possible then to rigorously show that since $$ \frac{a^t-e^{-y^2}}{a e^{y^2/t}-1}\to \frac{e^{-y^2}}{1-a},\qquad t\to+\infty $$ (because $a<1$) then $$ \sum_{j=1}^t\frac {a^{t-j}}{\sqrt j} =\frac{ t^{-1/2}}{1-a}(1+o(1)),\qquad t\to+\infty. $$