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

The Correct Option is A

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

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

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

= \(∫^1_0 \frac{1}{(1+x^2)(1+x)}dx\)

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

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

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

\(=\frac{π}{8}+\frac{1}{4}\;In2\)

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