Waste less time on Facebook — follow Brilliant.
×

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
1 year, 10 months ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears 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} \( \boxed{123} \)

Comments

Sort by:

Top Newest

What makes you think it can be simplified?

Pi Han Goh - 1 year, 10 months ago

Log in to reply

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 - 1 year, 10 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...