# Find the limit: lim_{nrightarrowinfty}(frac{1}{n+1}+frac{1}{n+2}+...+frac{1}{2n})

Question
Limits and continuity
Find the limit:
$$\lim_{n\rightarrow\infty}(\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n})$$

2020-10-28
Consider the given limit,
$$\lim_{n\rightarrow\infty}(\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n})$$
now write the limit in the summation form,
$$\lim_{n\rightarrow\infty}(\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n})=\lim_{n\rightarrow\infty}\sum_{r=1}^n(\frac{1}{n+r})$$
$$=\lim_{n\rightarrow\infty}\sum_{r=1}^n\frac{1}{n}\cdot(\frac{1}{1+\frac{r}{n}})$$
now, convert it into definite integral as follows,
substitute $$\frac{r}{n}=x\Rightarrow dx=\frac{1}{n}$$
so the limits of integration will be
$$r\rightarrow1,x\rightarrow0$$
$$r\rightarrow n,x\rightarrow1$$
so,
$$\lim_{n\rightarrow\infty}\sum_{r=1}^n\frac{1}{n}\cdot(\frac{1}{1+\frac{r}{n}})=\int_0^1\frac{1}{1+x}dx$$
$$=[\ln(1+x)]_0^1$$
$$=(\ln(2)-\ln(1))$$
$$=\ln(2)$$
hence the value of the limit will be ln(2).

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