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

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SUMMARY

The discussion centers on proving the equation pi/4 - ln(sqrt(2)) = 1 - 1/2 - 1/3 + 1/4 + 1/5... using the power series for arctan. The arctan series is defined as arctan(x) = 1 - (1/3)x^3 + (1/5)x^5 + ... with the general term being ((-1)^n)(x^(2n+1))/(2n+1). The user identifies that arctan(1) equals pi/4 and explores the series expansion, seeking clarification on the convergence of the series involving -1/2 + 1/4 - 1/6... and its relation to logarithmic properties, particularly log(sqrt(2)) = (1/2)log(2).

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  • Knowledge of the arctangent function and its series representation
  • Basic calculus concepts, including convergence of infinite series
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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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