Question

$\sum _{n=1}^{\infty }\frac{1}{(n+1)(n+2)(n+3)....(n+k)}$ is equal to

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

$T_n=\frac{1}{(n+1)(n+2)(n+3)....(n+k)}$

$V_{n-1}=\frac{1}{(n+1)(n+2)(n+3)....(n+k-2)(n+k-1)}$

$V_n-V_{n-1}=\frac{1}{(n+2)(n+3)....(n+k-1)}[\frac{1}{n+k}-\frac{1}{n+1}]$

$=\frac{1-k}{(n+1)(n+2)(n+3)....(n+k-1)(n+k)}$

$V_n-V_{n-1}=(1-k)T_n$

$\therefore (1-k)T_n=V_n-V_{n-1}$

$(1-k)T_1=V_1-V_0$

$(1-k)T_2=V_2-V_1$

$(1-k)T_3=V_3-V_2$

$...........................$

$(1-k)T_n=V_n-V_{n-1}$

$-------------------$

Adding $1-k)S_n=V_n-V_0$

or$(1-k)S_n=\frac{1}{(n+1)(n+2)......(n+k)}-\frac{1}{\mathrm{1.2.3....}k}(1-k{)}_{n\to \infty }^{Lt}{S}_n=0-\frac{1}{k!}$

${}_{n\to \infty }^{Lt}{S}_n=\frac{1}{(k-1)k!}$