Question

The value of $\underset{n\to \infty }{lim}n[\frac{1}{(n+1)(n+2)}+\frac{1}{(n+2)(n+4)}+\cdots +\frac{1}{6n^2}]=\log k$, then $2k=$

Answer 1

$L=\underset{n\to \infty }{lim}\sum _{r=1}^{n}n\cdot \frac{1}{(n+r)(n+2r)}\Rightarrow L=\underset{n\to \infty }{lim}\frac{1}{n}\sum _{r=1}^n\frac{1}{(1+r/n)(1+2r/n)}\Rightarrow L=\underset{0}{\overset{1}{\int }}\frac{dx}{(1+x)(1+2x)}\Rightarrow L=\underset{0}{\overset{1}{\int }}(\frac{-1}{1+x}+\frac{2}{1+2x})dx\Rightarrow L={[-\log (1+x)+\log (1+2x\left)\right]}_0^1\Rightarrow L=\log \left(3/2\right)\Rightarrow 2k=3$