$\lim _{n \rightarrow \infty}\left(\frac{n^2}{\left(n^2+1\right)(n+1)}+\frac{n^2}{\left(n^2+4\right)(n+2)}+\frac{n^2}{\left(n^2+9\right)(n+3)}+\ldots+\frac{n^2}{\left(n^2+n^2\right)(n+n)}\right)$ is equal to
Select the correct option:
A
$\frac{\pi}{8}+\frac{1}{4} \log _e 2$
B
$\frac{\pi}{4}+\frac{1}{8} \log _e 2$
C
$\frac{\pi}{4}-\frac{1}{8} \log _e 2$
D
$\frac{\pi}{8}+\log _e \sqrt{2}$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$\begin{aligned} & \lim _{n \rightarrow \infty}\left(\sum_{r=1}^n \frac{n^2}{\left(n^2+r^2\right)(n+r)}\right) \\ & =\lim _{n \rightarrow \infty}\left(\sum_{r=1}^n \frac{1}{n\left(1+\left(\frac{r}{n}\right)^2\right)\left(1+\left(\frac{r}{n}\right)\right)}\right) \\ & =\int_0^1 \frac{d x}{\left(1+x^2\right)(1+x)}=\frac{1}{2} \int_0^1 \frac{1-x}{1+x^2} d x+\frac{1}{2} \int_0^1 \frac{1}{1+x} d x \\ & =\frac{1}{2} \int\left(\frac{1}{1+x^2}-\frac{x}{1+x^2}\right) d x+\frac{1}{2}(\ln (1+x))_0^1 \\ & =\frac{1}{2}\left[\tan { }^{-1} x-\frac{1}{2} \ell n\left(1+x^2\right)\right]_0^1+\frac{1}{2} \ell n 2 \\ & =\frac{1}{2}\left[\frac{\pi}{4}-\frac{1}{2} \ell n 2\right]+\frac{1}{2} \ell n 2 \\ & =\frac{\pi}{8}+\frac{1}{4} \ell n 2\end{aligned}$
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