$\lim _{n \rightarrow \infty}\left(\frac{1}{1+n}+\frac{1}{2+n}+\frac{1}{3+n}+\ldots+\frac{1}{2 n}\right)$ is equal to :-

(C) $\log _{\mathrm{e}}\left(\frac{3}{2}\right)$

(D) $\log _{\mathrm{e}}\left(\frac{2}{3}\right)$

JEE Main · 2023 · 01 Feb 2023 · Shift 1 · Limits

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

is equal to :-

Answer: (B)

The sum

\sum_{r = 1}^{n} \frac{1}{n + r}

can be written as

\frac{1}{n} \sum_{r = 1}^{n} \frac{1}{1 + \frac{r}{n}}

, which is a Riemann sum for

f (x) = \frac{1}{1 + x}

[0, 1]

. As

n \to \infty

it tends to

\int_{0}^{1} \frac{d x}{1 + x} = [lo g_{e} (1 + x)]_{0}^{1}

. Evaluating between the limits gives the value of the limit as

lo g_{e} 2

JEE Main · 2023 · 01 Feb 2023 · Shift 1 · Single correct option

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