Mathematics - Limits Question with Solution | TestHub

The solution is given by:

$S_n=c-\frac{1}{3(n+1)(n+2)(n+3)}$

For $n=1$:

$$S_1=c-\frac{1}{3\cdot 2\cdot 3\cdot 4}$$

Given that $S_1=\frac{1}{1\cdot 2\cdot 3\cdot 4}$, we can find $c$:

$$\frac{1}{1\cdot 2\cdot 3\cdot 4}=c-\frac{1}{3\cdot 2\cdot 3\cdot 4}$$

$$c=\frac{1}{24}+\frac{1}{72}=\frac{3+1}{72}=\frac{4}{72}=\frac{1}{18}$$

As $n\to \infty$, $S_n\to c$, so $S_n\to \frac{1}{18}$.