# Riemann subtotal of $e$ on partition $Z = (0, \frac{1}{2^n}, \frac{2}{2^n},.., 1)$

Given the intermediate points $$\alpha = (\frac{1}{2^n}, \frac{2}{2^n},...,1)$$ and $$y=\frac{1}{2^n}$$. Prove that, the subtotal of $$e$$ on $$(Z, \alpha)$$ equals:

$$e^{y}\frac{e - 1}{\frac{e^y - 1}{y}} = e^yy\frac{e - 1}{e^y - 1}$$

This is currently my approach. The subtotal of $$e$$ on $$(Z, \alpha)$$ can be calculated directly like following:

$$\sum_{i = 1}^{n}(\frac{i}{2^n} - \frac{i - 1}{2^n})e^{\frac{i}{2^n}} = y\sum_{i = 1}^{n}e^{\frac{i}{2^n}} = e^yy \sum_{i = 1}^{n}e^{i} = e^yy (\frac{1 - e^{n + 1}}{1 - e} - 1) = e^yy (\frac{e - e^{n + 1}}{1 - e})$$

Hence the following equality must be true:

$$\frac{(e - 1)}{e^y - 1} = \frac{e-e^{n + 1}}{1 - e}$$

I can not go further by this point and would appreciate hints.