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Find $$S=x-\frac{2x^3}{3}+\frac{x^5}{5}+\frac{x^7}{7}-\frac{2x^9}{9}+\frac{x^{11}}{11}+\cdots=\sum_{n=0}^\infty\frac{x^{6n+1}}{6n+1}-\frac{2x^{6n+3}}{6n+3}+\frac{x^{6n+5}}{6n+5}$$

I could not find Regular pattern but I tried to differentiate $S$ as: $$\frac{dS}{dx}=1-2x^2+x^4+x^6-2x^8+x^{10}+\cdots$$ any clue here?

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Umesh shankar
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    What is the context for where this came from? Without additional details there is nothing stopping the next term in the sequence that you didn't write to be $\frac{x^{12}}{\sqrt{\pi e}}$. That is not to say that I would expect such a sequence with that as the next term to occur naturally, but from what we have seen we have no way of definitively knowing how often the extra factor of $-2$ occurs. Does it occur specifically with exponents of $x$ a multiple of $3$? Specifically at exponents of $x$ a power of $3$? – JMoravitz Sep 20 '17 at 04:04

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$$M=\frac x1+\frac{x^3}3+\frac{x^5}5+\frac{x^7}7+\frac{x^9}9+\frac{x^{11}}{11}+\cdots$$ is the Maclaurin series of $\tanh^{-1}x$ for $|x|<1$. The difference between this and $S$ is $$M-S=\frac{3x^3}3+\frac{3x^9}9+\frac{3x^{15}}{15}+\dots$$ $$=\frac{(x^3)^1}1+\frac{(x^3)^3}3+\frac{(x^3)^5}5+\dots=\tanh^{-1}x^3$$ Thus $$S=\tanh^{-1}x-\tanh^{-1}x^3,\ |x|<1$$

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