Infinite sum of reciprocal products

The given sum can be written as:

$\begin{aligned} S & = \frac 14 + \frac 2{4 \times 7} + \frac 3{4 \times 7 \times 10} + \frac 4{4 \times 7 \times 10 \times 13} + \cdots \\ & = \frac 14 + \frac 6{3 \times 4 \times 7} + \frac 9{3 \times 4 \times 7 \times 10} + \frac {12}{3 \times 4 \times 7 \times 10 \times 13} + \cdots \\ & = \frac 14 + \frac {7-1}{3 \times 4 \times 7} + \frac {10-1}{3 \times 4 \times 7 \times 10} + \frac {13-1}{3 \times 4 \times 7 \times 10 \times 13} + \cdots \\ & = \frac 14 + \frac 1{3 \times 4} - \cancel{\frac 1{3 \times 4 \times 7}} + \cancel{\frac 1{3 \times 4 \times 7}} - \cancel{\frac 1{3 \times 4 \times 7 \times 10}} + \cancel{\frac 1{3 \times 4 \times 7 \times 10}} - \cancel{\frac 1{3 \times 4 \times 7 \times 10 \times 13}} + \cdots \\ & = \frac 14 + \frac 1{3 \times 4} \\ & = \frac 13 \end{aligned}$

Hence, $\boxed{a=3}$

$\begin{aligned} S & = \lim_{m \to \infty} \sum_{n=1}^m \frac n{\prod_{k=1}^n (3k+1)} \\ & = \lim_{m \to \infty} \sum_{n=1}^m \frac n{3^n \prod_{k=1}^n \left(k+\frac 13\right)} \\ & = \lim_{m \to \infty} \sum_{n=1}^m \frac {n \Gamma \left(\frac 43 \right)}{3^n \Gamma \left(n+\frac 43\right)} \\ & = \lim_{m \to \infty} \frac {3^{-m-2}\left(-3m\Gamma \left(\frac 73 \right)+4 \times 3^m \Gamma \left(m+\frac73 \right) - 4\Gamma \left(\frac 73 \right)\right) \Gamma \left(\frac 43 \right)}{\Gamma \left(\frac 73 \right) \Gamma \left(m+\frac 73 \right)} \\ & = \frac {4\Gamma \left(\frac 43 \right)}{9\Gamma \left(\frac 73 \right)} = \frac {4\Gamma \left(\frac 43 \right)}{9 \times \frac 43 \Gamma \left(\frac 43 \right)} = \frac 13 \end{aligned}$

$\implies a = \boxed{3}$