Definite Integration Question 2
Question 2
- $\lim _{n \rightarrow \infty} \frac{n}{n^{2}+1^{2}}+\frac{n}{n^{2}+2^{2}}+\frac{n}{n^{2}+3^{2}}+\ldots+\frac{1}{5 n}$ is equal to (a) $\tan ^{-1}(3)$ (b) $\tan ^{-1}(2)$ (c) $\pi / 4$ (d) $\pi / 2$
(2019 Main, 12 Jan II)
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Answer:
Correct Answer: 2. $(a, d)$
Solution:
- Clearly,
$$ \begin{aligned} & \lim _{n \rightarrow \infty} \frac{n}{n^{2}+1^{2}}+\frac{n}{n^{2}+2^{2}}+\frac{n}{n^{2}+3^{2}}+\ldots+\frac{1}{5 n} \ &=\lim _{n \rightarrow \infty} \frac{n}{n^{2}+1^{2}}+\frac{n}{n^{2}+2^{2}}+\frac{n}{n^{2}+3^{2}}+\ldots+\frac{n}{n^{2}+(2 n)^{2}} \ &=\lim {n \rightarrow \infty} \sum{r=1}^{2 n} \frac{n}{n^{2}+r^{2}} \end{aligned} $$
$$ \begin{aligned} & =\lim {n \rightarrow \infty} \sum{r=1}^{2 n} \frac{1}{1+\frac{r}{n}{ }^{2}} \cdot \frac{1}{n}=\int_{0}^{2} \frac{d x}{1+x^{2}} \ & \because \lim {n \rightarrow \infty} \sum{r=1}^{p n} \frac{1}{n} f \frac{r}{n}=\int_{0}^{p} f(x) d x \ & =\left[\tan ^{-1} x\right]_{0}^{2}=\tan ^{-1} 2 \end{aligned} $$
