Help Me

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

#Algebra

Easy Math Editor

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Comments

What makes you think it can be simplified?

Pi Han Goh - 5 years, 4 months 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

$2015-0.9228386+0.4586751\approx2014.5358365$

Kay Xspre - 5 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$