0

\begin{equation} \begin{split} \sum_{j=1}^{p}\left[\frac{1}{\frac{1}{p-\alpha(j-1)}(1-\frac{1}{p-\alpha(j-1)})^{p-j}} \cdot\frac{1}{(\theta-\frac{1}{p-\alpha(j-1)})(p-j+1)}\right] \\ \end{split} \end{equation} $p$ is a finite constant, $\theta$ is a constant between 0 to 1, $\alpha$ is a constant making $\theta-\frac{1}{p-\alpha(j-1)}$ bigger than 0.

Please help me calculate this sum, or find the upper and lower bound(especially the upper bound).

Thank you very much!

Zhen Cao
  • 9
  • Seems as though the question would be far more readable is $p_{n}N$ were just written as a single character, say $p$, since it is simply a constant? Also, it is an integer, yes? In addition to being finite. – phdmba7of12 Sep 06 '17 at 14:32
  • yes, it is a constant, an integer and finite. Thank you. – Zhen Cao Sep 07 '17 at 08:13
  • $\sum_{j=1}^{p}\left[\frac{1}{\frac{1}{p-\alpha(j-1)}(1-\frac{1}{p-\alpha(j-1)})^{p-j}} \cdot\frac{1}{(\theta-\frac{1}{p-\alpha(j-1)})(p-j+1)}\right]$ – phdmba7of12 Sep 07 '17 at 14:29

0 Answers0