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
 A: As was mentioned in the comments, the integral $$\int_0^1\frac{dx}{1-x^2}$$
diverges, so you cannot split the integral $$\int_0^1\left(\frac{1}{1-x^2}-\frac{2x}{1-x^4}\right)dx$$
up into parts.
A: Another similar approach: consider the Beta integral
$$\int_0^1 (1-x)^2 x^n d x = B(3, n+1) = \frac{\Gamma(3)\Gamma(n+1)}{\Gamma(3 + n+1)} = \frac{2! \cdot n!}{(n+3)!} = \frac{2}{(n+1)(n+2)(n+3)}$$
We get
$$\sum_{n\ge 0} \frac{1}{(4n+1)(4n+2)(4n+3)} = \frac{1}{2} \int_{0}^1 (1-x)^2 (\sum_{n\ge 0}x^{4n}) dx=\\ = \frac{1}{2} \int_0^1 \frac{(1-x)^2}{1 - x^4} d x = \frac{\log 2}{4}$$
A: Try this.
$$
F(x) := \sum_{i=0}^\infty\frac{x^{4i+3}}{(4i+1)(4i+2)(4i+3)},\quad |x| \le 1
$$
Problem: compute $F(1)$.
Differentiate, to get geometric series
$$
F'''(x) = \sum_{i=0}^\infty x^{4i} = \frac{1}{1-x^4},\quad |x|<1
$$
Now integrate this three times, using $F(0) = F'(0) = F''(0)=0$.
$$
F(x) = \frac{x^2-1}{4}\arctan x + \frac{x^2}{4}\operatorname{atanh} x
+\frac{1}{8}\log(1+x) - \frac{1}{8}\log(1-x)
+\frac{x}{4}\log(1-x^2)-\frac{x}{4}\log(1+x^2)
$$
and then take limit
$$
\lim_{x\to 1^-}F(x) = \frac{\log 2}{4}
$$
A: 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 $S=0.1723286 \ldots$
A: Thinking about generalized harmonic numbers
$$\frac{1}{(4 i+1) (4 i+2) (4 i+3)}=\frac{1}{2 (4 i+1)}+\frac{1}{2 (4 i+3)}-\frac{1}{(4i+2)}$$
$$S_n=\sum_{i=0}^n\frac{1}{(4 i+1) (4 i+2) (4 i+3)}$$
$$\sum_{i=0}^n \frac{1}{ (4 i+1)}=\frac{1}{4} \left(\psi \left(n+\frac{5}{4}\right)-\psi  \left(\frac{1}{4}\right)\right)$$
$$\sum_{i=0}^n \frac{1}{ (4 i+3)}=\frac{1}{4} \left(\psi \left(n+\frac{7}{4}\right)-\psi
   \left(\frac{3}{4}\right)\right)$$
$$\sum_{i=0}^n \frac{1}{ (4 i+2)}=\frac{1}{4} \left(\psi \left(n+\frac{3}{2}\right)-\psi  \left(\frac{1}{2}\right)\right)$$
$$S_n=\frac{1}{4} \left(H_{2 n+\frac{3}{2}}-H_{n+\frac{1}{2}}\right)$$ Using the asymptotics, then
$$S_n=\frac{\log (2)}{4}-\frac{1}{128 n^2}+O\left(\frac{1}{n^3}\right)$$
Thinking about the gaussian hypergeometric function
$$\sum_{i=0}^\infty\frac{x^{2i}}{(4 i+1) (4 i+2) (4 i+3)}=\frac{3 x \, _2F_1\left(\frac{1}{4},1;\frac{5}{4};x^2\right)+x \,
   _2F_1\left(\frac{3}{4},1;\frac{7}{4};x^2\right)-3 \tanh ^{-1}(x)}{6 x}$$ and, if $x \to 1$, the limit.
Just, for the fun, notice that for $x=0$, the result is already $\frac 16=0.1667$ not "so far" from $\frac{\log(2)}4=0.1733$.
A: Find the limit
$$\frac{1}{1\cdot2\cdot3}+\frac{1}{5\cdot6\cdot7}+\frac{1}{9\cdot10\cdot11}+\cdots$$
The lim can be writen as:
$$S=\lim_{n\rightarrow\infty}\sum_{k=0}^n\frac{1}{(4k+1)(4k+2)(4h+3)}$$
$$=\frac{1}{2}\sum_{k=0}^{\infty}\frac{1}{4k+2}\left(\frac{1}{4k+1}-\frac{1}{4k+3}\right)=\frac{1}{2}\sum_{k=0}^{\infty}\left(\frac{1}{(4k+1)(4k+2)}-\frac{1}{(4k+3)(4k+3)}\right)$$
$$=\frac{1}{2}\sum_{k=0}^{\infty}\left\{\left(\frac{1}{4k+1}-\frac{1}{4k+2}\right)-\left(\frac{1}{4k+2}-\frac{1}{4k+3}\right)\right\}=\frac{1}{2}(S_1-S_2)$$
Now we will use the special function (digamma function):
$$\Psi(x)=\frac{d}{dx}\ln\Gamma(x)$$
and the identity
$$\sum_{k=0}^{\infty}\left(\frac{1}{k+y}-\frac{1}{k+x}\right)=\Psi(x)-\Psi(y)$$
With $y=\frac{1}{4},\quad x=\frac{2}{4}$:
$$\sum_{k=0}^{\infty}\left(\frac{4}{4k+1}-\frac{4}{2k+2}\right)=4\sum_{k=0}^{\infty}\left(\frac{1}{4k+1}-\frac{1}{2k+2}\right)=\left(\Psi\left(\frac{2}{4}\right)-\Psi\left(\frac{1}{4}\right)\right)\rightarrow$$
$$S_1=\frac{1}{4}\left(\Psi\left(\frac{2}{4}\right)-\Psi\left(\frac{1}{4}\right)\right)$$
With $y=\frac{2}{4},\quad x=\frac{3}{4}$:
$$\sum_{k=0}^{\infty}\left(\frac{4}{4k+2}-\frac{4}{2k+3}\right)=4\sum_{k=0}^{\infty}\left(\frac{1}{4k+2}-\frac{1}{2k+3}\right)=\left(\Psi\left(\frac{3}{4}\right)-\Psi\left(\frac{2}{4}\right)\right)\rightarrow$$
$$S_2=\frac{1}{4}\left(\Psi\left(\frac{3}{4}\right)-\Psi\left(\frac{2}{4}\right)\right)$$
Hence
$$S=-\frac{1}{8}\left(\Psi\left(\frac{1}{4}\right)-2\Psi\left(\frac{1}{2}\right)+\Psi\left(\frac{3}{4}\right)\right)$$
The particular values of $\Psi(x)$ are:
$\psi\left(\frac{1}{2}\right)=-C-2\ln2$
$\psi\left(\frac{1}{4}\right)=-C-\frac{\pi}{2}-3\ln2$
$\psi\left(\frac{3}{4}\right)=-C+\frac{\pi}{2}-3\ln2$
Whwre $C$ is the Euler's constante.
Replacing in $S$ we obtine
$$S=\frac{\ln2}{4}$$
