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Let the sum be $S$. First of all, it is easy to see that $$S=\frac{1}{2}\sum_{n=-\infty}^\infty f(n),\quad\mbox{where}\quad f(z)=\frac{1}{(6z+1)(6z+5)}.$$ This is true because $f(z)=f(-1-z)$. Then by the summation formula (see any undergraduate Complex Variables textbook) $$\sum_{n=-\infty}^\infty f(n)=-\sum_a{\mathrm{res}}_a f(z)\pi\cot\pi z,$$ where summation is over all poles of $f$, that is $a_1=-1/6$ and $a_2=-5/6$. Computing the residues (each of them equals $\pi\sqrt{3}/24$) we obtain the result.

Let the sum be $S$. First of all, it is easy to see that $$S=\frac{1}{2}\sum_{n=-\infty}^\infty f(n),\quad\mbox{where}\quad f(z)=\frac{1}{(6z+1)(6z+5)}.$$ This is true because $f(z)=f(-1-z)$. Then by the summation formula (see any undergraduate Complex Variables textbook) $$\sum_{n=-\infty}^\infty f(n)=-\sum_a{\mathrm{res}}_a \left(f(z)\pi\cot\pi z\right),$$ where summation is over all poles of $f$, that is $a_1=-1/6$ and $a_2=-5/6$. Computing the residues (each of them equals $-\pi\sqrt{3}/24$) we obtain the result.

Let the sum be $S$. First of all, it is easy to see that $$S=\frac{1}{2}\sum_{n=-\infty}^\infty f(n),\quad\mbox{where}\quad f(z)=\frac{1}{(6z+1)(6z+5)}.$$ By the summation formula (see any undergraduate Complex Variables textbook) $$\sum_{n=-\infty}^\infty f(n)=-\sum_a{\mathrm{res}}_a f(z)\pi\cot\pi z,$$ where summation is over all poles of $f$, that is $a_1=-1/6$ and $a_2=-5/6$. Computing the residues (each of them equals $\pi\sqrt{3}/24$) we obtain the result.

Let the sum be $S$. First of all, it is easy to see that $$S=\frac{1}{2}\sum_{n=-\infty}^\infty f(n),\quad\mbox{where}\quad f(z)=\frac{1}{(6z+1)(6z+5)}.$$ This is true because $f(z)=f(-1-z)$. Then by the summation formula (see any undergraduate Complex Variables textbook) $$\sum_{n=-\infty}^\infty f(n)=-\sum_a{\mathrm{res}}_a f(z)\pi\cot\pi z,$$ where summation is over all poles of $f$, that is $a_1=-1/6$ and $a_2=-5/6$. Computing the residues (each of them equals $\pi\sqrt{3}/24$) we obtain the result.

Let the sum be $S$. First of all, it is easy to see that $$S=\frac{1}{2}\sum_{n=-\infty}^\infty f(n),\quad\mbox{where}\quad f(z)=\frac{1}{(6z+1)(6z+5)}.$$ By the summation formula (see ay undergraduate Complex Variables textbook) $$\sum_{n=-\infty}^\infty f(n)=-\sum_a{\mathrm{res}}_a f(z)\pi\cot\pi z,$$ where summation is over all poles of $f$, that is $a_1=-1/6$ and $a_2=-5/6$. Computing the residues (each of them equals $\pi\sqrt{3}/24$) we obtain the result.

Let the sum be $S$. First of all, it is easy to see that $$S=\frac{1}{2}\sum_{n=-\infty}^\infty f(n),\quad\mbox{where}\quad f(z)=\frac{1}{(6z+1)(6z+5)}.$$ By the summation formula (see any undergraduate Complex Variables textbook) $$\sum_{n=-\infty}^\infty f(n)=-\sum_a{\mathrm{res}}_a f(z)\pi\cot\pi z,$$ where summation is over all poles of $f$, that is $a_1=-1/6$ and $a_2=-5/6$. Computing the residues (each of them equals $\pi\sqrt{3}/24$) we obtain the result.