Calculate the infinite sum
$$\dfrac{1}{2\cdot 3\cdot 4}+ \dfrac{1}{4\cdot 5\cdot 6}+\dfrac{1}{6\cdot 7\cdot 8}+\cdots$$
I know this series is convergent by Comparison Test, but I can't understand how can I get the value of the sum.
Is there any easy way to calculate this?
Please someone help.
Rewrite the sum $\sum_{i=1}^{\infty }\frac {1}{(2i)(2i+1)(2i+2)}$ as
$$ \sum_{i=1}^{\infty }\frac {(2i+1)-2i}{(2i)(2i+1)(2i+2)} = \sum_{i=1}^{\infty }\frac {1}{(2i)(2i+2)} -\sum_{i=1}^{\infty }\frac {1}{(2i+1)(2i+2)} $$
Or using partial fractions
$$ \frac{1}{4} \sum_{i=1}^{\infty} \left(\frac{1}{i} - \frac{1}{i+1}\right) - \sum_{i=1}^{\infty} \left(\frac{1}{2i+1} - \frac{1}{2i+2}\right) $$
The left sum telescopes to $1$, so we get that $ \frac{1}{4} \sum_{i=1}^{\infty} \left(\frac{1}{i} - \frac{1}{i+1}\right) = \frac14$
For the right sum
$$\sum_{i=1}^{\infty} \left(\frac{1}{2i+1} - \frac{1}{2i+2}\right) = \frac13 - \frac14 + \frac15-\frac16 + \dots $$
We use series expansion for $\ln(1+x)$
$$ \ln(1+x) = \sum_{k=1}^{n} \frac{(-1)^{k-1}x^k}{k} = x-\frac{x^2}{2}+\frac{x^3}{3} - \frac{x^4}{4} + \dots $$
Plug in $x=1$, to get that
$$ \ln 2 = 1 - \frac12 + \frac13 - \frac14 + \frac15 - \dots $$
Or
$$\frac13 - \frac14 + \frac15-\frac16 + \dots = \ln 2 - \frac12 $$
and that's our right sum so the final sum is equal to
$\frac14 - \ln2 + \frac12 = \frac34-\ln2 $
Hint. First observe that $$ \frac{1}{2i(2i+1)(2i+2)}=\frac{1}{2}\left(\frac{1}{2i(2i+1)}-\frac{1}{(2i+1)(2i+2)}\right)=\frac{1}{2}\left(\frac{1}{2i}-\frac{2}{2i+1}+\frac{1}{2i+2}\right) $$ Then $$ \sum_{i=1}^n\frac{1}{2i(2i+1)(2i+2)}=\sum_{i=1}^n\left(\frac{1}{2i}-\frac{1}{2i+1}\right)-\frac{1}{4}+\frac{1}{2(2n+2)}\\=\sum_{i=1}^n\frac{1}{2i(2i+1)}-\frac{1}{4}+\frac{1}{2n(2n+2)} $$ Clearly $$ \sum_{i=1}^n\frac{1}{2i(2i+1)}=\int_0^1\int_0^x(t+t^3+t^5+\cdots+t^{2n-1})\,dt\,dx\longrightarrow\int_0^1\int_0^x\frac{t\,dt}{1-t^2}\,dx\\=\frac{1}{2}\int_0^1\int_0^x\left(\frac{1}{1-t}-\frac{1}{1+t}\right)dt\,dx=-\frac{1}{2}\int_0^1\left(\log(1-x)+\log(1+x)\right)\,dx=\cdots $$ We have $$ \int_0^1 \log(1-x)\,dx=\int_0^1 \log x\,dx=\left.x\log x-x\right|_0^1=-1, \\ \int_0^1\log(1+x)\,dx=\int_1^2\log x\,dx=\left.x\log x-x\right|_1^2=2\log 2-1. $$ Hence $$ \sum_{i=1}^\infty\frac{1}{2i(2i+1)}=1-\log 2 $$ and finally $$ \sum_{i=1}^\infty\frac{1}{2i(2i+1)(2i+2)}=\frac{3}{4}-\log 2. $$
Let us use partial fraction decomposition and the identity
$$\frac{1}{n+1}=\int_0^1 x^n dx$$
to evaluate $$\begin{align}S&=\frac{1}{2·3·4}+\frac{1}{4·5·6}+\frac{1}{6·7·8}+...\\ \\ &=\sum_{k=0}^\infty \frac{1}{(2k+2)(2k+3)(2k+4)}=\sum_{k=0}^\infty \frac{1}{2} \left(\frac{1}{2k+2}-\frac{2}{2k+3}+\frac{1}{2k+4}\right)\\ \\ &=\frac{1}{2} \sum_{k=0}^\infty \int_0^1 \left(x^{2k+1}-2x^{2k+2}+x^{2k+3}\right)dx=\frac{1}{2} \int_0^1 (x-2x^2+x^3) \sum_{k=0}^\infty x^{2k}dx\\ \\ &=\frac{1}{2} \int_0^1 \frac{x(1-x)^2}{1-x^{2}}dx=\frac{1}{2} \int_0^1 \frac{x(1-x)}{1+x}dx\\ \\ &=\frac{1}{2} \int_0^1 \frac{-x^2+x}{1+x}dx=\frac{1}{2} \int_0^1 \frac{-x^2-x+2x+2-2}{1+x}dx\\ \\ &=\frac{1}{2} \int_0^1 \left(-x+2-\frac{2}{1+x}\right)dx=\frac{1}{2} \left(-\frac{x^2}{2}+2x-2\log(1+x)\right)|_0^1\\ \\ &=\frac{3}{4}-\log(2)\\ \end{align}$$
This series has all terms positive and thus is a direct proof of the inequality $$\log(2)<\frac{3}{4}$$
Similar series and integrals may be found in Series and integrals for inequalities and approximations to $\log(n)$.
In the style suggested by Michael Rozenberg's comment (solution 2), let us take one and two terms out of Mercator series
$$\begin{align} S_0=\log(2)&=\sum_{k=0}^\infty\left(\frac{1}{2k+1}-\frac{1}{2k+2}\right)\\ &=1-\sum_{k=0}^\infty \left(\frac{1}{2k+2}-\frac{1}{2k+3}\right)\\ &=1-\frac{1}{2}+\sum_{k=0}^\infty \left(\frac{1}{2k+3}-\frac{1}{2k+4}\right)\\ &=\frac{1}{2}+\sum_{k=0}^\infty \left(\frac{1}{2k+3}-\frac{1}{2k+4}\right)\\ \end{align}$$
to obtain
$$S_1 = \sum_{k=0}^\infty \left(\frac{1}{2k+2}-\frac{1}{2k+3}\right)=1-\log(2)$$
and $$S_2=\sum_{k=0}^\infty \left(\frac{1}{2k+3}-\frac{1}{2k+4}\right)=\log(2)-\frac{1}{2}$$
This reduces the evaluation to
$$\begin{align}S&=\sum_{k=0}^\infty \frac{1}{(2k+2)(2k+3)(2k+4)}\\ \\ &=\frac{1}{2}\sum_{k=0}^\infty\left(\frac{1}{2k+2}-\frac{2}{2k+3}+\frac{1}{2k+4}\right)\\ \\ &=\frac{1}{2}\sum_{k=0}^\infty\left(\frac{1}{2k+2}-\frac{2}{2k+3}+\frac{1}{2k+4}\right)\\ \\ &=\frac{1}{2}\sum_{k=0}^\infty\left(\frac{1}{2k+2}-\frac{1}{2k+3}-\frac{1}{2k+3}+\frac{1}{2k+4}\right)\\ \\ &=\frac{S_1-S_2}{2}=\dfrac{1-\log(2)-\left(\log(2)-\dfrac{1}{2}\right)}{2}=\dfrac{\dfrac{3}{2}-2\log(2)}{2}\\ \\ &=\frac{3}{4}-\log(2)\\ \end{align}$$
the semidifference of the precomputed series.
