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$\sum_{k=0}^\infty \frac{(-1)^k}{2k+1}$

How do I find the sum above, and I am confused that how to find the sums which are not geometric series or telescoping series.

I know Taylor series and Taylor expansion, but is it possible to reverse it from the series?

Jamie Chang
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2 Answers2

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You simply have the Taylor Series for $\arctan(x)$ at $x=1$ $$\arctan(x) = x-\frac{x^3}{3}+\frac{x^5}{5} ...$$ Put $x=1$, we get the required sum to be $\arctan(1) = \pi/4$

Hope this helps

Aditya Sriram
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Let $$f(x)=\sum_{k=0}^\infty\frac{(-1)^kx^{2k+1}}{2k+1}$$ We want $f(1)$. What is $df/dx $?

Empy2
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  • $\frac{df}{dx} = \frac{1-(-x^2)^\infty}{1+x^2}$

    It should be $\frac{df}{dx} = \frac{1}{1+x^2}$ if x<1

    But now x=1, could we really conclude $\frac{df}{dx} = \frac{1}{1+x^2}$ ?

     – Jamie Chang Apr 23 '20 at 13:30
  • The solution was already posted. However, I was wondering how to proceed form here. We obtain $f^{\prime}(x) = \sum_{k\geq 0} (-x^2)^k = 1/(1+x^2), x<1$. How do we conclude now? – Peter Wacken Apr 23 '20 at 13:30
  • The integral of $1/(1+x^2)$ is arctan. To make it rigorous, you might find the remainder term and show it approaches zero. – Empy2 Apr 23 '20 at 13:32