mathop {lim }limits_{n to infty } left[ {frac | vedclass

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(D) The given limit is $L = \mathop {\lim }\limits_{n 	o \infty } \sum_{r=0}^{n} \frac{1}{n+r}$.We can rewrite the expression as:$L = \mathop {\lim }

Vedclass · Accepted Answer

$\log_e 2$

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(D) The given limit is $L = \mathop {\lim }\limits_{n 	o \infty } \sum_{r=0}^{n} \frac{1}{n+r}$.We can rewrite the expression as:$L = \mathop {\lim }\limits_{n 	o \infty } \sum_{r=0}^{n} \frac{1}{n(1 + \frac{r}{n})} = \mathop {\lim }\limits_{n 	o \infty } \frac{1}{n} \sum_{r=0}^{n} \frac{1}{1 + \frac{r}{n}}$.Using the definition of a definite integral as the limit of a sum,$\mathop {\lim }\limits_{n 	o \infty } \frac{1}{n} \sum_{r=0}^{n} f(\frac{r}{n}) = \int_0^1 f(x) dx$,where $f(x) = \frac{1}{1+x}$.Thus,$L = \int_0^1 \frac{1}{1+x} dx$.Evaluating the integral:$L = [\log_e(1+x)]_0^1 = \log_e(1+1) - \log_e(1+0) = \log_e 2 - \log_e 1 = \log_e 2$.

$\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{1}{n} + \frac{1}{{n + 1}} + \frac{1}{{n + 2}} + \dots + \frac{1}{{2n}}} \right] = $

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