Mathematics

Evaluate: $$\underset {n \rightarrow \infty} \lim \displaystyle \sum_{r=0}^{n-1}\frac{1}{n+r}$$


ANSWER

$$\log 2$$


SOLUTION
$$\lim\limits_{n \to \infty} \displaystyle\sum_{r=0}^{n-1} \dfrac{1}{n+r}$$

$$=\lim\limits_{n \to \infty} \dfrac{1}{n} \displaystyle\sum_{r=0}^{n-1} \dfrac{1}{\left(1+\dfrac{r}{n}\right)}$$

Now, when $$ r=0, \dfrac{r}{n}= 0$$ 
and when $$ r=n-1, \lim\limits_{n \to \infty} \dfrac{r}{n}=1$$

Now, the given summation can be written as-
$$ \displaystyle\int_{0}^{1} \dfrac{1}{1+x} dx$$

$$=\left[\log{(1+x)}\right]_{0}^{1}$$

$$=\log{2}-\log{1}=\log2$$

Hence, the answer is option-(A).
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Single Correct Hard Published on 17th 09, 2020
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