SATHEE: Indefinite Integration 4 Question 6

Indefinite Integration 4 Question 6

6. Show that, $lim_{n \to \infty} \frac{1}{n + 1} + \frac{1}{n + 2} + \dots + \frac{1}{6 n} = \log 6$ .

$(1981, 2 M)$

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Solution:

$lim_{n \to \infty} \frac{1}{n + 1} + \frac{1}{n + 2} + \dots + \frac{1}{6 n} = \sum_{r = 1}^{5 n} \frac{1}{n + r}$

$\begin{aligned} = lim_{n \to \infty} \frac{1}{n} \sum_{r = 1}^{5 n} \frac{1}{1 + \frac{r}{n}} \\ = \int_{0}^{5} \frac{d x}{1 + x} = [\log (1 + x)]_{0}^{5} = \log 6 - \log 1 = \log 6 \end{aligned}$

Here,

Indefinite Integration 4 Question 6

6. Show that, $lim_{n \to \infty} \frac{1}{n + 1} + \frac{1}{n + 2} + \dots + \frac{1}{6 n} = \log 6$ .

Solution:

13

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Indefinite Integration 4 Question 6

6. Show that, limn→∞1n+1+1n+2+…+16n=log⁡6.

Solution:

13

Engineering Preparation

Medical Preparation

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6. Show that, $lim_{n \to \infty} \frac{1}{n + 1} + \frac{1}{n + 2} + \dots + \frac{1}{6 n} = \log 6$ .