# How to evaluate the following

#### bincybn

##### Member
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
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.