Evaluate : $S=\frac{1}{1\cdot2\cdot3}+\frac{1}{5\cdot6\cdot7}+\frac{1}{9\cdot10\cdot11}+\cdots$ Evaluate:$$S=\frac{1}{1\cdot2\cdot3}+\frac{1}{5\cdot6\cdot7} + \frac{1}{9\cdot10\cdot11}+\cdots$$to infinite terms
My Attempt:
The given series$$S=\sum_{i=0}^\infty \frac{1}{(4i+1)(4i+2)(4i+3)} =\sum_{i=0}^\infty \left(\frac{1}{2(4i+1)}-\frac{1}{4i+2}+\frac{1}{2(4i+3)}\right)=\frac{1}{2}\sum_{i=0}^\infty \int_0^1 \left(x^{4i}-2x^{4i+1}+x^{4i+2}\right) \, dx$$
So,$$S=\frac{1}{2}\int_{0}^{1}\left(\frac{1}{1-x^4}-\frac{2x}{1-x^4} + \frac{x^2}{1-x^4}\right)dx=\frac{1}{2} \int_0^1 \left(\frac{1+x^2}{1-x^4}-\frac{2x}{1-x^4}\right)\,dx = \frac{1}{2} \int_0^1 \left(\frac{1}{1-x^2}-\frac{2x}{1-x^4}\right)\,dx$$
$$=\frac{1}{2}\int_{0}^1\frac{1}{1-x^2}dx-\int_{0}^{1}\frac{2x}{1-x^4}dx=\frac{1}{2}\int_{0}^1\frac{1}{1-x^2}dx-\frac{1}{2}\int_{0}^{1}\frac{1}{1-y^2}dy=0(y=x^2)$$
which is obviously absurd since all terms of $S$ are positive.
But if I do like this then I am able to get the answer, $$S=\frac{1}{2}\int_{0}^{1}\left(\frac{1}{1-x^4}-\frac{2x}{1-x^4}+\frac{x^2}{1-x^4}\right)dx=\frac{1}{2}\int_{0}^{1}\frac{(1-x)^2}{1-x^4}dx=\frac{1}{2}\int_{0}^{1}\left(\frac{1}{1+x}-\frac{x}{1+x^2}\right)dx=\frac{\ln2}{4}$$
What is wrong with the previous approach

We have
$$
\begin{array}{l}
 S = \frac{1}{{1 \cdot 2 \cdot 3}} + \frac{1}{{5 \cdot 6 \cdot 7}} + \frac{1}{{9 \cdot 10 \cdot 11}} +  \cdots  =  \\ 
  = \sum\limits_{0 \le k} {\frac{1}{{\left( {4k + 1} \right)^{\,\overline {\;3\,} } }}}  = \sum\limits_{0 \le k} {\frac{{\Gamma \left( {4k + 4} \right)}}{{\Gamma \left( {4k + 1} \right)}}}
  = \sum\limits_{0 \le k} {t_{\,k} }  \\ 
 \end{array}
$$
where $ x^{\,\overline {\,k\,} } $ represents the Rising Factorial
Then we have
$$
\begin{array}{l}
 t_{\,0}  = \frac{1}{{1 \cdot 2 \cdot 3}} = \frac{1}{6} \\ 
 \frac{{t_{\,k + 1} }}{{t_{\,k} }} = \frac{{\Gamma \left( {4k + 5} \right)}}
{{\Gamma \left( {4k + 8} \right)}}\frac{{\Gamma \left( {4k + 4} \right)}}{{\Gamma \left( {4k + 1} \right)}}
 = \frac{{\left( {4k + 1} \right)^{\,\overline {\;4\,} } }}{{\left( {4k + 4} \right)^{\,\overline {\;4\,} } }}
 = \prod\limits_{j = 0}^3 {\frac{{\left( {4k + 1 + j} \right)}}{{\left( {4k + 4 + j} \right)}}}  =  \\ 
  = \prod\limits_{j = 0}^3 {\frac{{\left( {k + 1/4 + j/4} \right)}}{{\left( {k + 1 + j/4} \right)}}}
  = \prod\limits_{j = 0}^2 {\frac{{\left( {k + 1/4 + j/4} \right)}}{{\left( {k + 5/4 + j/4} \right)}}}  \\ 
 \end{array}
$$
that is the ratio of the consecutive addenda is a rational function of $k$.
So we can represent the sum as the Generalized Hypergeometric function
$$
S = \frac{1}{6}{}_4F_{\,3} \left( {\left. {\begin{array}{*{20}c}
   {1,\frac{1}{4},\;\frac{2}{4},\;\frac{3}{4}\;}  \\
   {1 + \frac{1}{4},\;1 + \frac{2}{4},\;1 + \frac{3}{4}}  \\
\end{array}\;} \right|\;1} \right)
$$
which gives $S=0.1723286 \ldots$
