Question

$lim _{ n → ∞}\bigg (\frac{n^2}{(n^2+1)(n+1)}+\frac{n^2}{(n^2+4)(n+2)}+\frac{n^2}{(n2+9)(n+3)}.....+ \frac{n^2}{(n^2+n^2)(n+n)}\bigg)$is equal to

• A. $\frac{π}{8}+\frac{1}{4} \;log_e2$
• B. $\frac{π}{4}+\frac{1}{8}\; log_e2$
• C. $\frac{π}{4}-\frac{1}{8}\; log_e2$
• D. $\frac{π}{8}+\frac{1}{8}\; log_e\sqrt2$

Answer 1

Answer: (A)

$\frac{π}{8}+\frac{1}{4} \;log_e2$

Answer 2

$lim _{ n → ∞}\bigg (\frac{n^2}{(n^2+1)(n+1)}+\frac{n^2}{(n^2+4)(n+2)}+\frac{n^2}{(n2+9)(n+3)}.....+ \frac{n^2}{(n^2+n^2)(n+n)}\bigg)$

= $lim _{n → ∞} ∑^n_{ r=1} \frac{n^2}{(n^2+r^2)(n+r)}$

= $lim_{ n → ∞} ∑^n_{ r=1} \frac{1}{\bigg[1+(\frac{r}{n})^2\bigg]\bigg[1+(\frac{r}{n})\bigg]}$

= $∫^1_0 \frac{1}{(1+x^2)(1+x)}dx$

= $\frac{1}{2} ∫^1_0\bigg [\frac{ 1}{1+x}-\frac{(x-1)}{(1+x^2)}\bigg]dx$

= $\frac{1}{2}\bigg[\;In(1+x)-\frac{1}{2}\;In(1+x^2)+tan^{-1}x\bigg]^1_0$

= $\frac{1}{2}\bigg[\frac{π}{4}+\frac{1}{2}\;In2\bigg]$

$=\frac{π}{8}+\frac{1}{4}\;In2$

Hence, the correct option is (A): $\frac{π}{8}+\frac{1}{4} \;log_e2$