The value of $\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=0}^{2 n-1} \frac{n^{2}}{n^{2}+4 r^{2}}$ is:

(D) $\frac{1}{2} \tan ^{-1}(4)$

(A) $\frac{1}{4} \tan ^{-1}(4)$

(B) $\frac{1}{2} \tan ^{-1}(2)$

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Limits

The value of

lim_{n \to \infty} \frac{1}{n} \sum_{r = 0}^{2 n - 1} \frac{n ^{2}}{n ^{2} + 4 r ^{2}}

is:

Answer: (D)

Rewrite the sum as a Riemann sum:

n \to \infty lim \frac{1}{n} r = 0 \sum 2 n - 1 \frac{n ^{2}}{n ^{2} + 4 r ^{2}} = n \to \infty lim r = 0 \sum 2 n - 1 \frac{1}{n} \cdot \frac{1}{1 + 4 ( r / n ) ^{2}}

Let

x = \frac{r}{n}

, then

Δ x = \frac{1}{n}

, and as

n \to \infty

, this becomes:

\int_{0}^{2} \frac{1}{1 + 4 x ^{2}} d x

Substitute

u = 2 x

d u = 2 d x

\int_{0}^{4} \frac{1}{1 + u ^{2}} \cdot \frac{d u}{2} = \frac{1}{2} [tan^{- 1} (u)]_{0}^{4} = \frac{1}{2} tan^{- 1} (4)

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Single correct option

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