TITLE: Evaluate : $S=\frac{1}{1\cdot2\cdot3}+\frac{1}{5\cdot6\cdot7}+\frac{1}{9\cdot10\cdot11}+\cdots$
QUESTION [9 upvotes]: 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

REPLY [2 votes]: 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
$$S=\frac{\ln2}{4}$$