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

JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Limits

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

is equal to :

Answer: (C)

Let

H_{n} = 1 + \frac{1}{2} + \dots + \frac{1}{n}

. By the Cauchy-Schwarz inequality

H_{n}^{2} \leq n \sum_{k = 1}^{n} \frac{1}{k ^{2}} < 2 n

, because

\sum_{k = 1}^{n} \frac{1}{k ^{2}} < 1 + \sum_{k = 2}^{n} \frac{1}{k ( k - 1 )} = 2 - \frac{1}{n}

. So

0 < \frac{H _{n}}{n} < \frac{2}{n}

, and using

ln (1 + t) < t

for

t > 0

0 < n ln (1 + \frac{H _{n}}{n ^{2}}) < n \cdot \frac{H _{n}}{n ^{2}} = \frac{H _{n}}{n} < \frac{2}{n} .

The bound tends to

0

, so the logarithm of the given expression tends to

0

and the limit is

1

JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Single correct option

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