limn→∞ 1n+1+1n+2+…+16n is equal to - Infinity Learn by Sri Chaitanya

A
$\log 2$
B
$\log 3$
C
$\log 5$
D
$\log 6$

Solution:

$\begin{array}{l} lim_{n \to \infty} (\frac{1}{n + 1} + \frac{1}{n + 2} + \dots + \frac{1}{6 n}) \\ = lim_{n \to \infty} (\frac{1}{n + 1} + \frac{1}{n + 2} + \dots + \frac{1}{n + 5 n}) \\ = lim_{n \to \infty} \sum_{r = 1}^{5 n} (\frac{1}{n + r}) = lim_{n \to \infty} \frac{1}{n} \sum_{r = 1}^{5 n} (\frac{1}{1 + \frac{r}{n}}) \end{array}$

$∵ Lower limit of r = 1$

$∴ Lower limit of integration = lim_{n \to \infty} \frac{1}{n} = 0 ∵ Upper limit of r = 5 n ∴ Upper limit of integration = lim_{n \to \infty} \frac{5 n}{n} = 5 from eq (i) \int_{0}^{5} \frac{1}{1 + x} dx = [\log (1 + x)]_{0}^{5} = \log 6 - \log 1 = \log 6$

$lim_{n \to \infty} (\frac{1}{n + 1} + \frac{1}{n + 2} + \dots + \frac{1}{6 n})$ is equal to

Solution:

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