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

Question
Limits and continuity
asked 2020-10-27
Find the limit:
\(\lim_{n\rightarrow\infty}(\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n})\)

Answers (1)

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).
0

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