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Do you know how to calculate $$\lim_{n\to\infty}\left(\dfrac1{n + 1} + \dfrac1{n + 2} + \cdots + \dfrac1{n + n}\right)\,?$$

Would be it correct to present this as $$\lim_{n\to\infty}\left(\sum_{k = 1}^n\dfrac1{n + k}\right)\,?$$

paweta
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2 Answers2

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Would be it correct to present this as $$\lim_{n\to\infty}\left(\sum_{k = 1}^n\dfrac1{n + k}\right)\,?$$

Yes.

Do you know how to calculate

One way it to see that it is an integral.

$$\begin{align} \lim_{n\to\infty}\left(\sum_{k = 1}^n\dfrac1{n + k}\right) &= \lim_{n\to\infty}\left(\frac1n\sum_{k = 1}^n\dfrac1{1 + k/n}\right) \\ &= \int_0^1 \frac1{1+x}dx\\ &= \ln(1+x)\Big|_{x=0}^{x=1} = \ln2 - \ln1\\ &= \ln2 \approx 0.693 \end{align}$$ The integral is the limiting process of dividing interval $[0,1]$ in $n$ parts of width $1/n$ and considering the area under the function $f(x) =1/(1+x)$. That area is split into $n$ rectangles of heights $f(x)$ and same widths $1/n$. As $k$ ranges from $1$ to $n$, $k/n$ finally ranges from $0$ to $1$.

emacs drives me nuts
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  • but where this 1/n dissapears ? – paweta Feb 27 '20 at 17:26
  • or we can consider that 1/n like a infinitely small (dx in integral ) ? – paweta Feb 27 '20 at 17:29
  • @paweta : The $1/n$ does not disappear. In Riemann integral with equi-distant division of the area, the area under the function is divided in $n$ rectangles of height $f(x)$ and same width $1/n$. Loosely speaking, the $1/n$ becomes the $dx$ in the integral. – emacs drives me nuts Feb 27 '20 at 17:54
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HINT

The formula $$H_n=\ln n + \gamma + O\left(\frac1n\right)$$ is derived Apostol's Analytic number theory page 55.

You are looking at the limit of $H_{2n} - H_n$, can you finish this yourself?

gt6989b
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    Well, currently I'm studying calculus II, and I haven't seen this formula yet. Would it be easier way to solve it ? – paweta Feb 27 '20 at 16:56
  • @paweta see the answers in the duplicate question, converting to an integral. – gt6989b Feb 27 '20 at 17:00