That is a one complicated denominator

Relevant wiki: Telescoping Series - Sum

Note that

$(k+1)(k+2)-k(k+3)=2 \\ \frac{1}{k(k+3)}-\frac{1}{(k+1)(k+2)}=\frac{2}{k(k+1)(k+2)(k+3)} \\ \frac{1}{2}\left(\frac{1}{k(k+3)}-\frac{1}{(k+1)(k+2)}\right)=\frac{1}{k(k+1)(k+2)(k+3)}$

Given expression divided by $81$ is equivalent to

$\lim_{n\rightarrow\infty}\sum_{k=1}^{n} \frac{1}{k(k+1)(k+2)(k+3)} \\ =\lim_{n\rightarrow\infty}\sum_{k=1}^{n} \frac{1}{2}\left(\frac{1}{k(k+3)}-\frac{1}{(k+1)(k+2)}\right) \\ =\lim_{n\rightarrow\infty}\sum_{k=1}^{n} \frac{1}{2}\left\{\frac{1}{3}\left(\frac{1}{k}-\frac{1}{k+3}\right)-\left(\frac{1}{k+1}-\frac{1}{k+2}\right)\right\} \\ =\frac{1}{6}\lim_{n\rightarrow\infty}\sum_{k=1}^{n} \left(\frac{1}{k}-\frac{1}{k+3}\right)- \frac{1}{2}\lim_{n\rightarrow\infty}\sum_{k=1}^{n} \left(\frac{1}{k+1}-\frac{1}{k+2}\right) \\ =\frac{1}{6}\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}\right)-\frac{1}{2}\times\frac{1}{2} \\ =\frac{11}{36}-\frac{1}{4} \\ =\frac{1}{18}$

Since we divided the original expression by $81$ , it's time we multiplied it back.

$81\times\frac{1}{18}=\frac{9}{2}=\boxed{4.5}$

Just a description of how to solve: 1) Partial fraction decomposition 2) Make use of telescoping property of resulting series

I solve in this way Let the kth term of the series be 1/((k)(k+1)(k+2)(k+3)) Here the trick is following So we have to apply partial fraction for the two factors only taking the largest and smallest factors . First Do the partial fraction of 1/((k)(k+3)) Note take the lower and Highest factors from the given factors.Here I took k and k+3. So we have 3/((k)(k+3)) =1/k - 1/(k+3) Hence 1/((k)(k+3) = (1/3){1/k -1/(k+3)} And Hence By Multiplying the above equation by 1/((k+1)(k+2 ) ) we have 1/((k)(k+3)(k+1)(k+2) =(1/3){1/((k)(k+1)(k+2)) - 1/((k+3)(k+1)(k+2))} Now Put k=1 ,2 ,3 ....and add telescopically We will get the Following Sum to infinity of the given Series will be = 81* (1/3){1/((1)(2)(3) - 1/((k+3)(k+1)(k+2))} Where k tends to infinity. As Here rest of the term gets cancelled out. =(81/3){1/((1)(2)(3) - 0} As the second term is Zero when k tends to infinity. So Sum to the infinity of the given Series is =81/((1)(2)(3)(3))=9/2=4.5 Hence Answer is 4.5