That's too much!

$\sum _{ k=1 }^{ \infty }{ \frac { 1 }{ k\left( 1+12k \right) } } \\ =\quad \frac { 1 }{ 12 } \sum _{ k=1 }^{ \infty }{ \frac { 12 }{ k\left( 1+12k \right) } } \\ =\quad \frac { 1 }{ 12 } \sum _{ k=1 }^{ \infty }{ \frac { 1 }{ k\left( \frac { 1 }{ 12 } +k \right) } } ={ H }_{ \frac { 1 }{ 12 } }\\ =12-3\left( ln2+\frac { ln3 }{ 2 } \right) -\pi \left( 1+\frac { \sqrt { 3 } }{ 2 } \right) +2\sqrt { 3 } ln\left( \sqrt { 2-\sqrt { 3 } } \right)$

$\displaystyle\sum_{r=1}^{n} \frac{1}{r+a} = \psi(n+a+1)-\psi(a+1)$

Split the given fraction as $\frac{1}{k(1+12k)} = \frac{1}{k} - \frac{1}{k+\frac{1}{12}}$

Taking summations upto $n$ both the sides you get

$\displaystyle \sum_{k=1}^{n} \frac{1}{k} - \frac{1}{k+\frac{1}{12}}= \psi(n+1) - \psi(1) - \psi(n + \frac{13}{12}) + \psi(\frac{13}{12})$

using $\psi(1) = -\gamma$

and

Take limits $n$ to $\infty$ both the sides we get...

$\displaystyle\boxed{\psi(\frac{13}{12}) + \gamma}$