ISI2015-MMA-26

Question

$\displaystyle{}\underset{n \to \infty}{\lim} \frac{1}{n} \bigg( \frac{n}{n+1} + \frac{n}{n+2} + \cdots + \frac{n}{2n} \bigg)$ is equal to

• A. $\infty$
• B. $0$
• C. $\log_e 2$
• D. $1$

Comments

Is $0$ is the answer?

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No. Option C is the correct answer.

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why on slving i m getting

1 – 1/2n

puttting n= infinite im getting 1

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You should show your method of the solution first, then only I would be able to comment further. Anyways, you can see the solution below.

Answer 1

Answer: C

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$$\begin{align*}\lim_{n\to\infty}\frac{1}{n}\left(\frac{n}{n+1}+\frac{n}{n+2}+\cdots+\frac{n}{2n}\right)=&\lim_{n\to\infty}\sum_{r=1}^n\frac{1}{n}\left(\frac{n}{n+r}\right) \\=&\lim_{n\to\infty}\sum_{r=1}^n\frac{1}{n}\left(\frac{1}{1+\frac{r}{n}}\right)\\ =& \int_0^1\frac{1}{1+x}dx\\ =& \log_e2\end{align*}$$

 

Here, $\frac{r}{n}=x$ and $\frac{1}{n}=dx$.