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I was reading a solution to a problem given in my book the problem was $$\lim_{n \to \infty}\left( \frac{1}{n + 1}+ \frac{1}{n + 2} + \frac{1}{n + 3 }+........+\frac{1}{6n}\right)$$ now it was then converted to $$ \lim_{n \to \infty}\sum_{r=1}^{5n}\left(\frac{1}{n+r} \right) =\lim_{n \to \infty}\frac{1}{n} \sum_{r=1}^{5n}\left(\frac{1}{1+r/n} \right)$$ i somewhat understand the conversion but what i don't get was that why summation is done till $ 5n$ not $6n$ and finally it was written $\int_{0}^{5}\frac{1}{1+x}dx$ i still somewhat get it buy why limits are from $0$ to $5$ while in summation it's from $1$ to $5n$ please don't give me external links and please explain me here

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In the sum, $\frac{1}{n}$ is analogous to $dx$, while $\frac{r}{n}$ is analogous to $x$. $r$ goes from $1$ to $5n$, so $\frac{r}{n}$ goes from $\frac{1}{n}$ to $5$. As $n\to\infty$, $\frac{1}{n}\to 0$, so as $n\to\infty$, $\frac{r}{n}$ goes from $0$ to $5$, hence $x$ goes from $0$ to $5$.

The summation goes to $5n$ because $n+r=6n$ if $r=5n$.

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  • $\begingroup$ that explained the concept behind it very well $\endgroup$ – Tesla Aug 21 '13 at 8:02
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The general term is $ \dfrac 1 {n+r} $. The last term is $\dfrac 1 {6n} = \dfrac 1 {n+5n} $. So the last $r$ is $5n$. Hence the summation is done till $5n$.

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