Hi I am doing some work in interacting particle systems.
I have this sum $$ \sum_{k=0}^{\infty} r^{\frac{k}{2}(2m-1-k)} $$ where $m$ is some integer and $r>1$ is real. I don't how to work this sum out.
If anyone knows what $$ \frac{1}{(-r^{m},\frac{1}{r})_{\infty}}\sum_{k=0}^{\infty} (r^{m-1})^{k}(\frac{1}{r})^{\frac{(k-1)k}{2}} $$
is where $\frac{1}{(-r^{m},\frac{1}{r})_{\infty}}$ is the q pochhammer symbol. This would be equivalently useful to solving the 1st series. Im thinking something from this document could help https://faculty.math.illinois.edu/~berndt/articles/q.pdf
Estimation of the $S=\sum\limits_{k=0}^{\infty} r^{\frac{k}{2}(2m-1-k)}$
(1) $S=r^{\frac{t^2}{2}}\sum\limits_{k=0}^{\infty} e^{-\frac{1}{2}(k-t)^2 lnr}$ where $t=\frac{2m-1}{2}$
Using that $f(x)=r^{\frac{t^2}{2}} e^{-\frac{1}{2}(x-t)^2 ln{r}} $ not increasing function on $[0,\infty)$ the following estimation is valid:
(2) $\int\limits_0^n f(x) dx +f(n)-f(0)\le r^{\frac{t^2}{2}}\sum\limits_{k=0}^{\infty} e^{-\frac{1}{2}(k-t)^2 lnr}\le \int\limits_0^n f(x) dx$
$f(n)= r^{\frac{t^2}{2}}e^{-\frac{1}{2}(n-t)^2 ln{r}}=e^{-\frac{1}{2}\big((n-t)^2-t^2\big) ln {r} }$ and $f(0)= r^{\frac{t^2}{2}}e^{-\frac{1}{2}t^2 lnr}=1$
if $n\rightarrow\infty$ then $f(n)\rightarrow 0 $
We get that:
(3) $\int\limits_0^\infty f(x) dx -1\le r^{\frac{t^2}{2}}\sum\limits_{k=0}^{\infty} e^{-\frac{1}{2}(k-t)^2 lnr}\le \int\limits_0^\infty f(x) dx$
(4) Dealing with $I=\int\limits_0^\infty f(x) dx=r^{\frac{t^2}{2}}\int\limits_0^\infty e^{-\frac{1}{2}(x-t)^2 lnr}dx$
Applying the $y=\frac{\sqrt{\ln{r}}}{\sqrt{2}}(x-t)$ substitution:
$I=r^{\frac{t^2}{2}}\big({\frac{2}{ln{r}}}\big)^{\frac{1}{2}}\int\limits_{-\frac{t\sqrt{ln{r}}}{\sqrt{2}}}^\infty e^{-y^2}dy=r^{\frac{t^2}{2}}\big({\frac{\pi}{2ln{r}}}\big)^{\frac{1}{2}}\big(erf(\infty)-erf(-\frac{t \sqrt{ln{r}}}{\sqrt{2}})\big)$
As $erf(\infty)=1$ and $erf(-\frac{t \sqrt{ln{r}}}{\sqrt{2}}\big)=-erf(\frac{t \sqrt{ln{r}}}{\sqrt{2}}\big)$
$I=r^{\frac{t^2}{2}}\sqrt{\frac{\pi}{2ln{r}}}\big(1+erf(t\sqrt{\frac{ln{r}}{2}}\big)$
If $t\sqrt{\frac{ln{r}}{2}}\gt\gt 1$ then $erf(t\sqrt{\frac{ln{r}}{2}})\rightarrow 1$ and $I\rightarrow 2r^{\frac{t^2}{2}}\sqrt{\frac{\pi}{2ln{r}}}$
Finally go back to (3)
$I-1\le S \le I$
If $I\gt\gt 1$ then $S\approx I$