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Proving $$\sum_{n=1}^{\infty }\frac{0.5^n}{(n)(n+1)(n+2)(n+3)}=\frac{5}{36}-\frac{\log(2)}{6}$$

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$$\sum_{n=1}^{\infty }\frac{0.5^n}{(n)(n+1)(n+2)(n+3)}\\= \sum_{n=1}^{\infty }0.5^n\left(\frac{1/6}n+\frac{-1/2}{n+1}+\frac{1/2}{n+2}+\frac{-1/6}{n+3}\right)\\ =\frac16\sum_{n=1}^{\infty }\frac{0.5^n}n-\frac1{2\cdot0.5}\sum_{n=2}^{\infty }\frac{0.5^n}n+\frac1{2\cdot0.5^2}\sum_{n=3}^{\infty }\frac{0.5^n}n-\frac1{6\cdot0.5^3}\sum_{n=4}^{\infty }\frac{0.5^n}n\\ =\frac16(\ln2)-(\ln2-0.5)+2(\ln2-0.5-0.5^2/2)-\frac43(\ln2-0.5-0.5^2/2-0.5^3/3)\\ =\frac5{36}-\frac16\ln2$$

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Another approach. From: $$\frac{1}{n(n+1)(n+2)(n+3)}=\frac{1}{6}\frac{\Gamma(n)\Gamma(4)}{\Gamma(n+4)}=\frac{1}{6}\int_{0}^{1} u^{n-1}(1-u)^3\,du $$ it follows that: $$\sum_{n\geq 1}\frac{1}{2^n n(n+1)(n+2)(n+3)}=\frac{1}{6}\int_{0}^{1}\frac{(1-u)^3}{2-u}\,du $$ and: $$ \int_{0}^{1}\frac{(1-u)^3}{2-u}\,du = \int_{0}^{1}\frac{x^3}{1+x}\,du=-\log 2+\int_{0}^{1}(1-x+x^2)\,dx=-\log 2+\frac{5}{6}, $$ hence: $$\sum_{n\geq 1}\frac{1}{2^n n(n+1)(n+2)(n+3)}=\color{red}{-\frac{\log 2}{6}+\frac{5}{36}}$$ as wanted.

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$$\sum\frac{x^n}{n(n+1)(n+2)(n+3)}=\sum\frac{x^n}6\left(\frac1n-\frac3{n+1}+\frac3{n+2}-\frac1{n+3}\right)$$

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This is not fully sketched answer. (It takes some time to type all the steps, sorry :-) )

See a similar question here. Using the same techniques I used there (explained briefly in the comments below my answer) we find that

$$\sum_{n=1}^{\infty}\frac{x^n}{n(n+1)(n+2)(n+3)}$$

is the Taylor expansion of $$f(x)=\frac{(6-18x+18x^2-6x^3)\ln(1-x)+6x-15x^2+11x^3}{36x^3}$$

thus $$\sum_{n=1}^{\infty}\frac{(\frac12)^n}{n(n+1)(n+2)(n+3)}=f(\frac12)=\frac{5}{36}-\frac16\ln2.$$

(Note that this answer is in essence similar to the suggestion by ADG.)

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