Question

Find the sum of infinite series $\frac{1+3}{1!}\left(lo{g}_{e}3\right)+\frac{1+3^2}{2!}(lo{g}_{e}3{)}^2+\frac{1+3^3}{3!}(lo{g}_{e}3{)}^3+...$

Answer 1

Let $S=\frac{(1+3)}{1!}\left({\log }_{e}3\right)+\frac{(1+3^2)}{2!}({\log }_{e}3{)}^2+\frac{(1+3^3)}{3!}({\log }_{e}3{)}^3+...$
$=(1+\frac{{\log }_{e}3}{1}+\frac{({\log }_{e}3{)}^2}{2!}+\frac{({\log }_{e}3{)}^3}{2!}+.....)+(1+\frac{3{\log }_{e}3}{1}+\frac{(3{\log }_{e}3{)}^2}{2!}+\frac{(3{\log }_{e}3{)}^3}{2!}+.....)-2$
$=e^{{\log }_{e}3}+e^{3{\log }_{e}3}-2$
$=3+27-2=28$