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I came across this limit evaluation method in my notebook, which is very poorly explained.

$$1+\frac{1}{2}+\cdots+\frac{1}{n}=\gamma+\mathcal{E}_n+ln(n)\space,\ \mathcal{E}_n\longrightarrow 0\space\ when\space\ n\longrightarrow\infty$$

Its application on one problem is the following

$$\text{Problem:}\space \lim_{n\to\infty} \frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}$$

$$\text{Solution:}\space\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}=\color{navy}{1+\frac{1}{2}+\cdots+\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}\cdots+\frac{1}{2n}-\left(1+\frac{1}{2}+\cdots+\frac{1}{n}\right)}=\gamma+\mathcal{E}_{2n}+ln(2n)-(\gamma+\mathcal{E}_n+ln(n))=\mathcal{E}_{2n}-\mathcal{E}_n+ln(2)$$

$$\text {And thereby,}$$

$$\lim_{n\to\infty} \frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}=\lim_{n\to\infty}\underbrace{\mathcal{E}_{2n}}_{\rightarrow 0}-\underbrace{\mathcal{E}_n}_{\rightarrow 0}+ln(2)=ln(2)$$

Alright, I see that the sequence is being manipulated in order to substitute it with the upward formula. What I fail to grasp is the navy highlighted step. The other thing I don't understand is the formula and what does it represent, why does it work at all.

Could someone elaborate on what is happening here ?

The formula is introduced in my Analysis course before the derivatives, integrals etc..

Thanks

powerline
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    The difference step is pretty straightforward, reduce the first n terms of the harmonic sum and you are left with the required sum. About $\gamma$ (Euler-Mascheroni) constant, you should read https://brilliant.org/wiki/euler-mascheroni-constant/ – LHF Jan 27 '20 at 05:32
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    The highlighted step is: sum from $1$ to $\frac{1}{2n}$ is the sum from $\frac{1}{n+1}$ to $\frac{1}{2n}$ plus the sum from $1$ to $\frac{1}{n}$ – fGDu94 Jan 27 '20 at 05:44
  • @Atticus I see it now. I read the article. Thanks for the clarification! What is $\mathcal {E}_n$ supposed to be in this case ? Is it maybe the epsilon neighborhood from the limit of a sequence definition ? – powerline Jan 27 '20 at 05:44
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    $\mathcal{E}_n$ is just a very small positive term (which is about $\frac{1}{2n}$) and can be negliged for large $n$ (as they do in the proof when it takes the limit $n\to \infty$) – LHF Jan 27 '20 at 05:49
  • @Atticus Alright, but wherefrom is the equation coming? I was thinking maybe it was derived from the $$\gamma=\lim_{n \to \infty}1+\frac{1}{2}+\cdots+\frac{1}{n}-ln(n) \space \implies |1+\frac{1}{2}+\cdots+\frac{1}{n}-ln(n)-\gamma|< \epsilon \space \iff 1+\frac{1}{2}+\cdots+\frac{1}{n}<\gamma+\epsilon+ln(n)$$. And that $\mathcal {E}_n$ is some $\epsilon$ which will eventually become $0$ for some $n$ big enough. – powerline Jan 27 '20 at 06:15
  • @GrigoriPerelman, you are correct. $H_{n}-\ln n$ is a strictly decreasing sequence, so your $\epsilon$ will become very small for large $n$. – LHF Jan 27 '20 at 06:17
  • @Atticus but I was just thinking, I formed an inequality from the definition where the RHS is strictly larger than LHS but in my question it actually says they are equal. What happens then that makes them equal ? – powerline Jan 27 '20 at 06:19
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    @GrigoriPerelman, It's just that $H_n - \ln n$ is a strictly decreasing sequence that converges toward $\gamma$. So we can say that $H_n-\ln n = \gamma + \mathcal{E}_n$, where $\mathcal{E}_n$ gets smaller every time $n$ increases. – LHF Jan 27 '20 at 06:22
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    @Atticus Got it! Thanks a lot man. – powerline Jan 27 '20 at 06:27
  • @Atticus I posted a problem solution using this equation asking for verification, you can check it out if you are interested. regards – powerline Jan 27 '20 at 06:55

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