$1-\frac{1}{5\cdot 3^2}-\frac{1}{7\cdot 3^3}+\frac{1}{11\cdot 3^5}+\frac{1}{13\cdot 3^6}--++\cdots.$ I 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.

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 $\frac{1}{2}\ln 7$.
