Question

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

• A. $\frac{1}{(k-1).(k-1)!}$
• B. $\frac{1}{k.k!}$
• C. $\frac{1}{(k-1).k!}$
• D. $\frac{1}{k!}$

Answer 1

Answer: (C)

$\frac{1}{(k-1).k!}$

The series:
$\frac{1}{\mathrm{2.3.4...}(k+1)}+\frac{1}{\mathrm{3.4.5...}(k+2)}+\frac{1}{\mathrm{4.5.6...}(k+3)}+...\infty =\frac{1}{k-1}.[\left(\frac{(k+1)-2}{\mathrm{2.3.4...}(k+1)}\right)+\left(\frac{(k+2)-3}{\mathrm{3.4.5...}(k+2)}\right)+...\infty ]=\frac{1}{k-1}.[(\frac{1}{\mathrm{2.3...}k}-\frac{1}{\mathrm{3.4...}(k+1)})+(\frac{1}{\mathrm{3.4...}(k+1)}-\frac{1}{\mathrm{4.5...}(k+2)})+...\infty ]=\frac{1}{k-1}.\left(\frac{1}{\mathrm{2.3...}k}\right)=\frac{1}{(k-1)k!}$
Hence, (C) is correct.