2007 TYS

Express

$\frac{2n+3}{n\left( n+1 \right)}$

in partial fractions and hence use the method of differences to find

${{\sum\limits_{n=1}^{N}{\frac{2n+3}{n\left( n+1 \right)}\left( \frac{1}{3} \right)}}^{n+1}}$

in terms of $N$.

[4]

Deduce the value of

$\sum\limits_{n=1}^{\infty }{\frac{2n+3}{n\left( n+1 \right)}{{\left( \frac{1}{3} \right)}^{n+1}}}$

[1]