Proving pi/4-ln(sqrt(2)) with Arctan Series

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Homework Help Overview

The discussion revolves around proving the equation pi/4 - ln(sqrt(2)) = 1 - 1/2 - 1/3 + 1/4 + 1/5... using the power series for arctan. The subject area includes series convergence and properties of logarithms.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the representation of arctan(1) and its relation to the series. There are attempts to relate the series to geometric series and the Taylor series for natural logarithms. Questions are raised about the equivalence of certain series and properties of logarithms.

Discussion Status

Participants are exploring various approaches to relate the series to the logarithmic expression. Some guidance has been offered regarding logarithmic properties, which appears to facilitate the discussion.

Contextual Notes

There is mention of difficulties in relating certain series and the need for hints, indicating that participants are working within the constraints of their understanding of series and logarithmic identities.

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Homework Statement



Using the power series for arctan show that pi/4-ln(sqrt(2))=1-1/2-1/3+1/4+1/5...


Homework Equations


arctan=1-(1/3)x^3+(1/5)x^5...(((-1)^n)(x^(2n+1)))/(2n+1)


The Attempt at a Solution



The first thing I noticed was that arctan(1)=pi/4 and represented 1-1/3+1/5...

I am having trouble figuring out what -1/2+1/4-1/6... is equivalent to besides a simple series.

If anyone has any hints. I would love to hear them because I am stuck.
 
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I have tried converting it back to a geometric series and comparing it to the natural log taylor series but haven't had any luck.
 
What's the sum from n=1 to infinity of \frac{\left(-1\right)^n}{n}? Also, do you see that log(\sqrt{2}) = \frac{1}{2}log(2)?
 
Thank you so much. It is always something simple. That log property makes this problem a lot easier.
 

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