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I want to evaluate the series $$1-\frac{1}{5\cdot 3^2}-\frac{1}{7\cdot 3^3}+\frac{1}{11\cdot 3^5}+\frac{1}{13\cdot 3^6}--++\cdots.$$ I can rewrite this as $$1+\sum_{n\geq 1} (-3)^{-3n}\left(\frac{3}{6n-1}+\frac{1}{6n+1}\right)$$ The answer should be $\frac{\ln7}{2}$. Although I can't see explicitly, since I get a log function, maybe it can be solved by differentiating some test function $f(x)$ and then substituting an appropriate number.

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  • $\begingroup$ I can confirm that this series adds to $\ln 7/2$ based on numerical calculations. $\endgroup$
    – subrosar
    Commented Oct 17, 2021 at 23:38

2 Answers 2

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For context and inspiration, let's remember that the linear factors in the power series for $\log(1+x)$ come from integrating a geometric series: $$ \log(1+x) = \int_0^x \frac1{1+t}\,dt = \int_0^x \sum_{n=0}^\infty (-1)^n t^n \,dt = \sum_{n=0}^\infty (-1)^n \frac{t^{n+1}}{n+1}. $$ So if we recognize our series as a special value of a power series, which turns out to be $$ f(x) = 1-\sqrt3\biggl( \frac{x^5}5 + \frac{x^7}7 - \frac{x^{11}}{11} - \frac{x^{13}}{13} + \frac{x^{17}}{17} + \frac{x^{19}}{19} - \frac{x^{23}}{23} - \frac{x^{25}}{25} + \cdots \biggr) $$ evaluated at $x=\frac1{\sqrt3}$, then we can work out what that power series is by writing it as an integral of a rational function: \begin{align*} f(x) &= 1 - \sqrt3 \int_0^x ( x^4 + x^6 - x^{10} - x^{12} + x^{16} + x^{18} - x^{22} - x^{24} + \cdots )\,dt \\ &= 1 - \sqrt3 \int_0^x (t^4+t^6)(1-t^6+t^{12}-t^{18}+\cdots)\,dt \\ &= 1 - \sqrt3 \int_0^x \frac{t^4+t^6}{1+t^6} \,dt = 1 - \sqrt3 \int_0^x \frac{t^4}{t^4-t^2+1} \,dt. \end{align*} Using partial fractions (well, telling my computer to use partial fractions), we obtain \begin{align*} f(x) &= 1 - \int_0^x \biggl( \sqrt3 + \frac{t-\sqrt3/2}{t^2-t\sqrt3+1} - \frac{t+\sqrt3/2}{t^2+t\sqrt3+1} \biggr) \,dt \\ &= 1 - x\sqrt3 - \frac{\log (x^2-x\sqrt3+1)}2 + \frac{\log (x^2+x\sqrt3+1)}2, \end{align*} and therefore the series we want is \begin{align*} f\biggl( \frac1{\sqrt3} \biggr) &= 1 - 1 - \frac{\log (1/3)}2 + \frac{\log (7/3)}2 = \frac{\log7}2 \end{align*} as predicted.

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  • $\begingroup$ Let aside I used both of the answers, I am thankful that you drew the motivation. (At any rate, because the calculation was easier, I chose the first answer. It was difficult for me to choose between easier computation and motivation.) $\endgroup$
    – Nugi
    Commented Oct 18, 2021 at 8:20
  • $\begingroup$ Yes, I realized later that it probably would make the calculation details easier if I had put some factors of 3 in my defining functions ... Rene Schipperus proved that correct :) $\endgroup$ Commented Oct 18, 2021 at 20:24
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Consider the series,

$$x-\frac{x^5}{5\cdot 3^2}-\frac{x^7}{7\cdot 3^3}+\frac{x^{11}}{11\cdot 3^5}+\frac{x^{13}}{13\cdot 3^6}--++\cdots.$$

Differentiate this and get $$1-\frac{x^4}{ 3^2}-\frac{x^6}{ 3^3}+\frac{x^{10}}{ 3^5}+\frac{x^{12}}{3^6}--++\cdots.$$

This is a geometric series, and sums to

$$3\frac{9-x^4}{27+x^6}$$ Integrating,

$$\int 3\frac{9-x^4}{27+x^6}dx=\frac{1}{2}(\ln(x^2+3x+3)- \ln(x^2-3x+3))+C$$

Note that $C=0$ by letting $x=0$ Therefore

$$x-\frac{x^5}{5\cdot 3^2}-\frac{x^7}{7\cdot 3^3}+\frac{x^{11}}{11\cdot 3^5}+\frac{x^{13}}{13\cdot 3^6}--++\cdots.$$ $$=\frac{1}{2}(\ln(x^2+3x+3)- \ln(x^2-3x+3))$$

Setting $x=1$ we get $\frac{1}{2}\ln 7$.

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