Mathematics - Limits Question with Solution | TestHub

Let $u_k=\frac{k+2}{k(k+1)(k+3)}$

$$\begin{array}{rl}& =\frac{(k+2{)}^2}{k(k+1)(k+2)(k+3)}\\ & =\frac{k^2+4k+4}{k(k+1)(k+2)(k+3)}\\ & =\frac{k(k+1)+3k+4}{k(k+1)(k+2)(k+3)}\\ & =\frac{1}{(k+2)(k+3)}+\frac{3}{(k+1)(k+2)(k+3)}+\frac{4}{k(k+1)(k+2)(k+3)}\end{array}$$

$$\begin{array}{rl}& =\left(\frac{1}{k+2}-\frac{1}{k+3}\right)-\frac{3}{2}\left[\frac{1}{(k+2)(k+3)}-\frac{1}{(k+1)(k+2)}\right]\\ & -\frac{4}{3}\left[\frac{1}{(k+1)(k+2)(k+3)}-\frac{1}{k(k+1)(k+2)}\right]\end{array}$$

Now, put $k=1,2,3,\dots ,n$ and add. Thus

$$\begin{array}{rl}{s}_u& =u_1+u_2+\dots .+u_n\\ & =\left(\frac{1}{3}-\frac{1}{n+3}\right)-\frac{3}{2}\left[\frac{1}{(n+2)(n+3)}-\frac{1}{2.3}\right]\\ & -\frac{4}{3}\left[\frac{1}{(n+1)(n+2)(n+3)}-\frac{1}{\mathrm{1.2.3}}\right]\end{array}$$

Therefore ${\lim }_{n\to \infty }{s}_n=\frac{1}{3}+\frac{3}{12}+\frac{4}{18}=\frac{29}{36}$