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

Find the limit:
$\underset{n\to \mathrm{\infty }}{lim}\left(\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}\right)$
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oppturf
Consider the given limit,
$\underset{n\to \mathrm{\infty }}{lim}\left(\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}\right)$
now write the limit in the summation form,
$\underset{n\to \mathrm{\infty }}{lim}\left(\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}\right)=\underset{n\to \mathrm{\infty }}{lim}\sum _{r=1}^{n}\left(\frac{1}{n+r}\right)$
$=\underset{n\to \mathrm{\infty }}{lim}\sum _{r=1}^{n}\frac{1}{n}\cdot \left(\frac{1}{1+\frac{r}{n}}\right)$
now, convert it into definite integral as follows,
substitute $\frac{r}{n}=x⇒dx=\frac{1}{n}$
so the limits of integration will be
$r\to 1,x\to 0$
$r\to n,x\to 1$
so,
$\underset{n\to \mathrm{\infty }}{lim}\sum _{r=1}^{n}\frac{1}{n}\cdot \left(\frac{1}{1+\frac{r}{n}}\right)={\int }_{0}^{1}\frac{1}{1+x}dx$
$=\left[\mathrm{ln}\left(1+x\right){\right]}_{0}^{1}$
$=\left(\mathrm{ln}\left(2\right)-\mathrm{ln}\left(1\right)\right)$
$=\mathrm{ln}\left(2\right)$
hence the value of the limit will be ln(2).
Jeffrey Jordon