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# Help Me

How to simplify this? $\frac {1} {e^{-2015} + 1} + \frac {1} {e^{-2014}+1} + \frac {1} {e^{-2013} +1 } + \cdots + \frac {1} {e^{-1} + 1}$

Note by Pandu Wb
2 years, 1 month ago

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What makes you think it can be simplified?

- 2 years, 1 month ago

The very furthest way to simplify this is to put it into the form of

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

then find the partial sum. I cannot elaborate the partial sum for $$\frac{1}{e^n+1}$$, but, according to Wolfram Alpha

$\sum_{n=1}^m \frac{1}{e^n+1} = \psi_{\frac{1}{e}}(1-i\pi)-\psi_{\frac{1}{e}}(1+m-i\pi)$

Where $$\psi_{n}$$ is a q-polygamma function, hence the partial sum is

$\sum_{n=1}^m (1-\frac{1}{e^n+1}) = m-\psi_{\frac{1}{e}}(1-i\pi)+\psi_{\frac{1}{e}}(1+m-i\pi)$

For $$m = 2015$$, this will be

$2015-0.9228386+0.4586751\approx2014.5358365$

- 2 years, 1 month ago