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How to evaluate this infinite series :$$S=1+\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{9}+\dfrac{1}{11}+....=1+\sum_{k=0}^{\infty}\frac{2(-1)^k}{(4k+3)(4k+5)}$$

therfore: $$S-1=\sum_{k=0}^{\infty}\frac{2(-1)^k}{(4k+3)(4k+5)}=\sum_{k=1}^{\infty}\frac{2(-1)^{k+1}}{(4k-1)(4k+1)}=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{(4k-1)}-\frac{(-1)^{k+1}}{(4k+1)}$$

Therfore: $$S-1=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{(4k-1)}-\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{(4k+1)}$$

How can I calculate this?

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    $\begingroup$ Use the identity $\frac1n=\int_0^1x^{n-1}~{\rm d}x$. This should convert it into a geometric series, which can be summed and then integrated. $\endgroup$ Commented May 15 at 12:58
  • $\begingroup$ You're allowed to separate a series into two subseries if the subseries both converge absolutely. However, in some series, commuting the terms changes the sum. What theorem allows separating this series into two subseries? $\endgroup$
    – nickalh
    Commented May 15 at 13:25
  • $\begingroup$ I have asked this question in $2019$, math.stackexchange.com/questions/3190908/… $\endgroup$ Commented May 15 at 13:38

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We have $$\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{(4k-1)}=\sum_{k=1}^{\infty}\int_0^1 (-1)^{k+1}x^{4k-2}\,dx=\int_0^1 \left(\sum_{k=1}^{\infty}(-1)^{k+1}x^{4k-2}\right)\,dx=\int_{0}^1\dfrac{x^2}{1+x^4}\,dx,$$

where the change of order between sum and integral is justified by the Dominated Convergence Theorem.

Analogously, $$\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{(4k+1)}=\sum_{k=1}^{\infty}\int_0^1 (-1)^{k+1}x^{4k}\,dx=\int_0^1 \left(\sum_{k=1}^{\infty}(-1)^{k+1}x^{4k}\right)\,dx=\int_{0}^1\dfrac{x^4}{1+x^4}\,dx,$$

Thus, we have $$S=1+\int_{0}^1\dfrac{x^2-x^4}{1+x^4}\,dx$$

An antiderivative of the integrand can be found by the standard methods (long division, partial fractions,etc) and is equal to $$\dfrac{\arctan\left(\frac{2x+\sqrt{2}}{\sqrt{2}}\right)}{\sqrt{2}}+\dfrac{\arctan\left(\frac{2x-\sqrt{2}}{\sqrt{2}}\right)}{\sqrt{2}}-x$$

Therefore, $$\int_{0}^1\dfrac{x^2-x^4}{1+x^4}\,dx=\dfrac {\arctan(1+\sqrt{2})-\arctan(1-\sqrt{2})}{\sqrt{2}}-1$$

Using that $\arctan(a)-\arctan(b)=\arctan\left(\dfrac{a-b}{1+ab}\right)$ we get $$\int_{0}^1\dfrac{x^2-x^4}{1+x^4}\,dx=\dfrac{\pi}{2\sqrt{2}}-1$$

from where it follows that $$S=1+\left(\dfrac{\pi}{2\sqrt{2}}-1\right)=\dfrac{\pi}{2\sqrt{2}}$$

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We could rewrite the summation as $\int_0^1 dx + \int_0^1 x^2 dx - \int_0^1 x^4 dx -\int_0^1 x^6 dx + ...$ This could be written as $\int_0^1 1+x^2-x^4-x^6+x^8+x^{10}+... dx$ This could be be simplified by writing these terms as a part of a GP. We could add and subtract 2 times all the negative parts of the integration. We get $$\int_0^1 1+x^2+x^4+x^6+... dx - (-2\int_0^1 x^4+x^6+x^{12}+x^{14}... dx)$$ And now we could write the first part of the integration as $$\int_0^1 \frac{1}{1-x^2} dx$$ and similarly we could simplify the final part of the summation by separating each power of consecutive two (sorry for such confusing language), so the final summation becomes $$\int_0^1 x^4+x^{12}+x^{20}+... dx + \int_0^1 x^6+x^{14}+x^{22}+... dx$$. This can now overall be simplified to $$\int_0^1 \frac{1}{1-x^2} dx + 2( \int_0^1 \frac{x^4}{1-x^8} dx + \int_0^1 \frac{x^6}{1-x^8} dx)$$. Now this question can be easily solved. Hope this helps.

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  • $\begingroup$ Hi, welcome to Math SE. Thanks for using MathJax to write your answer. Please note x^n only works as intended when the exponent is a single-character or a call to a backslashed method such as \frac{a}{b}. To get $x^{10}$, you need some curly braces viz.x^{10}. $\endgroup$
    – J.G.
    Commented May 15 at 13:25
  • $\begingroup$ The last integral is obviously a mistake, unless you meant 1/1. I'll leave it to you work out what you meant. perhaps \frac{1}{1-x^2} etc. $\endgroup$
    – nickalh
    Commented May 15 at 13:37

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