I want to find the value of $\displaystyle\sum_{n=1}^\infty \dfrac{\cos n}{n^22^n\binom{3n}{n}}.$
Since $\displaystyle\dfrac{1}{\binom{3n}{n}}=2n\beta(2n,n+1)=2n\int_0^1t^{2n-1}(1-t)^ndt$,
$\displaystyle\sum_{n=1}^\infty \dfrac{x^n}{n^2\binom{3n}{n}}=\sum_{n=1}^\infty \dfrac{2x^n}{n}\int_0^1t^{2n-1}(1-t)^ndt=\int_0^1\dfrac{2}{t}\left(\sum_{n=1}^\infty \dfrac{(x(1-t)t^2)^n}{n}\right)dt$.
Since $\displaystyle\sum_{n=1}^\infty\frac{z^n}{n}=-\ln{(1-z)}$, $\displaystyle\int_0^1\dfrac{2}{t}\left(\sum_{n=1}^\infty \dfrac{(x(1-t)t^2)^n}{n}\right)dt=-\int_0^1\dfrac{2}{t}\ln{\left(1-x(1-t)t^2\right)}dt$.
Therefore $\displaystyle\sum_{n=1}^\infty \dfrac{x^n}{n^2\binom{3n}{n}}=-\int_0^1\dfrac{2}{t}\ln{\left(1-x(1-t)t^2\right)}dt$.
If substitute $x=\dfrac{e^i}{2}$ and consider the real part , then $\displaystyle\sum_{n=1}^\infty \dfrac{\cos n}{n^22^n\binom{3n}{n}}=-\int_0^1\frac{\ln{\left(1-kt^2(1-t)+\frac{t^4(1-t)^2}{4}\right)}}{t}dt$, where $k=\cos1$.
Is it possible to find the exact value of this integral?
Thank in advances.