Mathematics - Limits Question with Solution | TestHub

The problem involves finding the limit of a series as $n$ approaches infinity. We use partial fraction decomposition.

Let $S_n={\sum }_{k=1}^n\frac{1}{k(k+1)(k+2)(k+3)}$.

We can rewrite the general term as:

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

This is a telescoping series. Thus,

$$S_n=\frac{1}{3}\left[\frac{1}{1\cdot 2\cdot 3}-\frac{1}{(n+1)(n+2)(n+3)}\right]$$

As $n\to \infty$, the second term approaches 0.

$$\underset{n\to \infty }{\lim }{S}_n=\frac{1}{3}\cdot \frac{1}{6}=\frac{1}{18}$$

Therefore, the answer is $\frac{1}{18}$.