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$a_n=\sum_{k=1}^{n} \frac{1}{n+k}=\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{2n}$

How to find $\lim a_n$?

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    $\begingroup$ Look at it as a Riemann sum. $\endgroup$ – Daniel Fischer Jan 4 '14 at 13:28
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With Eulero-Mascheroni : $$\sum_{k = 1}^{n}\frac{1}{k} - \log{n} \rightarrow \gamma$$ $$\sum_{k = 1}^{2n}\frac{1}{k} - \log{2n} \rightarrow \gamma$$ so $$\sum_{k = n+1}^{2n}\frac{1}{k} - \log{2n} +\log{n} \rightarrow 0$$ and then$$\sum_{k = n+1}^{2n}\frac{1}{k} \rightarrow \log{2}$$

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HINT

As Daniel Fisher suggested, rewrite 1/(n+k) as (1/n) 1/(1+k/n) and you are done through integration. I am sure you can take from here.


Below is a reformatted version of this answer

HINT

As Daniel Fisher suggested, rewrite $\dfrac 1{n+k}$ as $\dfrac {1}{n} \dfrac {1}{1+\frac k n}$ and you are done through integration. I am sure you can take from here.

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  • $\begingroup$ Do you mind if I type your answer in $\LaTeX$ below yours, keeping your version? $\endgroup$ – Git Gud Jan 4 '14 at 14:08
  • $\begingroup$ @GitGud. I am sorry for that (I even wrote it in my profile). I became almost blind a couple of years ago and I do not "see" what I type. Beside plain ASII, everything is very hard to me and I did not find a way to learn LaTex in my conditions. Be sure I really apologize. I thank you very much if you accept to LaTeXify for me. Cheers. $\endgroup$ – Claude Leibovici Jan 4 '14 at 14:28
  • $\begingroup$ I read your profile, I knew about your condition before I posted my comment. That's why I'm asking for your permission to add a $\LaTeX$ version of your answer (in your answer), while keeping the original version. $\endgroup$ – Git Gud Jan 4 '14 at 14:30
  • $\begingroup$ Again, thank you very much for doing it. $\endgroup$ – Claude Leibovici Jan 4 '14 at 14:31
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Brutally speaking, $$\sum_{k=1}^{n}\frac1{n+k}=\left(\sum_{k=1}^{2n}\frac1k\right)-\left(\sum_{k=1}^{n}\frac1k\right)\simeq\ln2n-\ln n=\ln\frac{2n}n=\ln2.$$ Rigourously, we have $$\sum_{k=1}^{n}\frac1{n+k}=\frac1n\cdot\sum_{k=1}^{n}\frac1{1+\frac kn}=\int_0^1\frac{dx}{1+x}=\ln(1+x)|_0^1=\ln2.$$

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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $$ \sum_{k = 1}^{n}{1 \over n + k} = \sum_{k = n}^{2n - 1}{1 \over k + 1} = \sum_{k = 0}^{2n - 1}{1 \over k + 1} - \sum_{k = 0}^{n - 1}{1 \over k + 1} =\bracks{\Psi\pars{2n + 1} - \Psi\pars{1}} - \bracks{\Psi\pars{n + 1} - \Psi\pars{1}} $$ $$ \sum_{k = 1}^{n}{1 \over n + k} = \Psi\pars{2n + 1} - \Psi\pars{n + 1} $$ Since $\Gamma\pars{z + 1} \sim \root{2\pi}z^{z + 1/2}\expo{-z}$ when $\verts{z} \gg 1$, $\Psi\pars{z + 1} \sim \ln\pars{z} + {1 \over 2z}$ $$ \sum_{k = 1}^{n}{1 \over n + k} \sim \bracks{\ln\pars{2n} + {1 \over 2\pars{2n}}} - \bracks{\ln\pars{n} + {1 \over 2n}} = \ln\pars{2} - {1 \over 4n} $$ $$\color{#0000ff}{\large% \lim_{n \to \infty}\sum_{k = 1}^{n}{1 \over n + k} = \ln\pars{2}} $$

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