$\lim_{n \rightarrow \infty}\left(\frac{n}{n^{2}+1^{2}}+\frac{n}{n^{2}+2^{2}}+\frac{n}{n^{2}+3^{2}}+\ldots+\frac{1}{5 n}\right)$ is equal to

JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Limits

lim_{n \to \infty} (\frac{n}{n ^{2} + 1 ^{2}} + \frac{n}{n ^{2} + 2 ^{2}} + \frac{n}{n ^{2} + 3 ^{2}} + \dots + \frac{1}{5 n})

is equal to

Answer: (D)

Each term has the form

\frac{n}{n ^{2} + r ^{2}} = \frac{1}{n} \cdot \frac{1}{1 + ( r / n ) ^{2}}

. The last term

\frac{1}{5 n} = \frac{n}{5 n ^{2}} = \frac{n}{n ^{2} + ( 2 n ) ^{2}}

corresponds to

r = 2 n

, so the sum runs over

r = 1

2 n

. As

n \to \infty

this is a Riemann sum with

Δ x = \frac{1}{n}

and

x = \frac{r}{n}

sweeping

[0, 2]

, so it becomes

\int_{0}^{2} \frac{d x}{1 + x ^{2}} = [tan^{- 1} x]_{0}^{2} = tan^{- 1} 2

JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Single correct option

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