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$$\lim_{n \rightarrow \infty} \sum^n_{k=1} \frac {1} {k+n} = \ln 2$$

I'm supposed to find the limit and get the right part as an answer. I don't know what to do with it. Could someone explain?

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marked as duplicate by Jyrki Lahtonen, Git Gud, Thomas Andrews, Lord_Farin, Norbert Jul 3 '13 at 20:01

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ I knnow this has been asked before. Can't find it, though. $\endgroup$ – Thomas Andrews Jul 3 '13 at 19:05
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Hint: Think of Riemann sums and note that $$ \sum\limits_{k=1}^n\frac{1}{k+n}=\sum\limits_{k=1}^n\frac{1}{1+\frac{k}{n}}\frac{1}{n} $$

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Write it as $$\lim_{n\rightarrow\infty}\sum_{k=1}^{n}\frac{1}{n}\frac{1}{\frac{k}{n}+1}$$

and look it as Riemann sums for the integral $$\int_{0}^{1}\frac{1}{x+1}dx$$

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Let me give another method without using the Riemann sum.

We use the asymptotic expansion $$\sum_{k=1}^n\frac{1}{k}=\log n+\gamma+o(1)$$ where $\gamma$ is the Euler's constant hence $$\sum_{k=1}^n\frac{1}{k+n}=\sum_{k=n+1}^{2n}\frac{1}{k}=\sum_{k=1}^{2n}\frac{1}{k}-\sum_{k=n}^{2n}\frac{1}{k}=\log(2n)-\log(n)+o(1)=\log 2+o(1)$$ and we deduce the desired result.

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  • $\begingroup$ +1 , though the "we know that" phrase may be far from true for someone asking such a basic question. $\endgroup$ – DonAntonio Jul 3 '13 at 19:34
  • $\begingroup$ You're right and thanks for the comment. $\endgroup$ – user63181 Jul 3 '13 at 19:36

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