Welcome to our community

Be a part of something great, join today!

How to evaluate the following

bincybn

Member
Apr 29, 2012
36
Hi all,

Can I get an upper bound of the below expression in terms of $\textbf{all } r,v,b$?

$\displaystyle \sum_{k=1}^{\infty}
\quad
\frac{r^{kb}}{\left(1-r^{k+1}\right)^{a}-\left(1-r^{k}\right)^{a}}
\quad : b>1
, 0<r,a<1$

Can we atleast obtain an upper bound for $\sum_{k=1}^{\infty}
\quad
\frac{1}{\left(1-r^{k+1}\right)^{a}-\left(1-r^{k}\right)^{a}}
$ ?


Please Note that $\sum_{k=1}^{\infty}
\quad
\left(1-r^{k+1}\right)^{a}-\left(1-r^{k}\right)^{a}
$ iteratively cancels up, leaving a single term. (Don't know if it is useful)

Kind regards,
bincy
 
Last edited:

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,790
Hi all,

Can I get an upper bound of the below expression in terms of $\textbf{all } r,v,b$?

$\displaystyle \sum_{k=1}^{\infty}
\quad
\frac{r^{kb}}{\left(1-r^{k+1}\right)^{a}-\left(1-r^{k}\right)^{a}}
\quad : b>1
, 0<r,a<1$

Can we atleast obtain an upper bound for $\sum_{k=1}^{\infty}
\quad
\frac{1}{\left(1-r^{k+1}\right)^{a}-\left(1-r^{k}\right)^{a}}
$ ?


Please Note that $\sum_{k=1}^{\infty}
\quad
\left(1-r^{k+1}\right)^{a}-\left(1-r^{k}\right)^{a}
$ iteratively cancels up, leaving a single term. (Don't know if it is useful)

Kind regards,
bincy
Hi bincybn! :)

Did you know that $(1+x)^a = 1 + \binom a 1 x + \binom a 2 x^2 + ...$?
This applies even if $a$ is a non-integer number, in which case $\binom a k$ is exactly what you would expect of it.
It's not quite an answer to your upper bound, but it is a step in its direction.