Question

For each positive integer n, let $s_n=\frac{3}{\mathrm{1.2.4}}+\frac{4}{\mathrm{2.3.5}}+\frac{5}{\mathrm{3.4.6}}+\dots \dots +\frac{n+2}{n(n+1)(n+3)}$. Then $\underset{n\to \infty }{lim}{s}_n$ equals

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

Let $u_k=\frac{k+2}{k(k+1)(k+3)}=\frac{(k+2{)}^2}{k(k+1)(k+2)(k+3)}=\frac{k^2+4k+4}{k(k+1)(k+2)(k+3)}=\frac{k(k+1)+3k+4}{k(k+1)(k+2)(k+3)}=\frac{1}{(k+2)(k+3)}+\frac{3}{(k+1)(k+2)(k+3)}+\frac{4}{k(k+1)(k+2)(k+3)}=(\frac{1}{k+2}-\frac{1}{k+3})-\frac{3}{2}[\frac{1}{(k+2)(k+3)}-\frac{1}{(k+1)(k+2)}]-\frac{4}{3}[\frac{1}{(k+1)(k+2)(k+3)}-\frac{1}{k(k+1)(k+2)}]Now,putk=1,2,3,\dots \dots ,nandadd.Thus{s}_u=u_1+u_2+\dots \dots +u_n=(\frac{1}{3}-\frac{1}{n+3})-\frac{3}{2}[\frac{1}{(n+2)(n+3)}-\frac{1}{2.3}]-\frac{4}{3}[\frac{1}{(n+1)(n+2)(n+3)}-\frac{1}{\mathrm{1.2.3}}]Therefore\underset{n\to \infty }{lim}{s}_n=\frac{1}{3}+\frac{3}{12}+\frac{4}{18}=\frac{29}{36}$