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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 1 year, 9 months ago

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What makes you think it can be simplified?

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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

\[2015-0.9228386+0.4586751\approx2014.5358365\]

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TopNewestWhat makes you think it can be simplified?

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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

\[2015-0.9228386+0.4586751\approx2014.5358365\]

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