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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

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

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