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Is $\lim\limits_{k\to\infty}\sum\limits_{n=k+1}^{2k}{\frac{1}{n}} = 0$?

I encountered the following sum in Boros & Moll's "Irresistible Integrals" question 5.2.11 (pg. 78):

$$\lim_{n \to \infty} \sum_{m=n}^{2n} \frac{1}{m}$$

How do I evaluate this sum to get a closed form?

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marked as duplicate by Marvis, J. M., Pedro Tamaroff, Rahul, Argon Aug 3 '12 at 22:29

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.

Take a look here as well: math.stackexchange.com/questions/102197/… –  Byron Schmuland Aug 3 '12 at 22:16

3 Answers 3

up vote 2 down vote accepted

If $H_x=1+\frac{1}{2}+\cdots+\frac{1}{x}$ (i.e. the $x^{th}$ Harmonic Number), we see

$$ \lim_{n \to \infty} \sum_{m=n}^{2n} \frac{1}{m}= \lim_{n \to \infty} \frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n}= \lim_{n \to \infty}\left(1+\frac{1}{2}+\cdots + \frac{1}{2n}\right)-\left(1+\frac{1}{2}+\cdots+\frac{1}{n-1}\right)= \lim_{n \to \infty} H_{2n}-H_{n-1} $$

You may recall that

$$\log x +\gamma+O(1)= H_x \quad (\text{as } x\to\infty)$$

Now, because $m \ge n \to \infty$ we may use $\log x$ instead of $H_x$. Then we have

$$ \lim_{n \to \infty} \sum_{m=n}^{2n} \frac{1}{m}= \lim_{n \to \infty} H_{2n}-H_{n-1}= \lim_{n \to \infty} (\log (2n)+\gamma)-(\log (n-1)+\gamma)= \lim_{n \to \infty} \log (2)+\log (n)-\log (n-1)= \log (2) $$

A quick numerical check seems to confirm this: If $n=1000$, the sum approximately equals $0.6938972430$ - three correct decimal places of $\log 2$.

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Great answer! This makes lots of sense to me. –  Heebie Aug 3 '12 at 22:15
The result is correct, but the argument is incomplete. You should justify that $H_{2n} - H_{n-1}$ has the same limit as $\log(2n) - \log(n-1)$. Notice that $H_n - \log n$ does not converge to $0$, so saying that $H_n \sim \log n$ isn't enough. –  Erick Wong Aug 3 '12 at 22:16
@Argon What about $H_n=\log(n)+\gamma+o(1)$ –  Pedro Tamaroff Aug 3 '12 at 22:20

Another way: $$\lim_{n\to\infty}\sum_{m=n}^{2n}\frac{1}{m}=\lim_{n\to\infty}\frac{1}{n}\sum_{m=n}^{2n}\frac{n}{m}=\int_{1/2}^1\frac{dx}{x}=\left.\log x\right|_{1/2}^1=-\log \frac{1}{2}=\log 2$$

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This one seems like the simplest answer. –  Michael Hardy Aug 3 '12 at 22:42
+1 for not needing the asymptotic expansion of $H_n$. –  Erick Wong Aug 3 '12 at 23:29

The Euler Maclaurin Sum Formula says that $$ \sum_{k=1}^n\frac1k=\log(n)+\gamma+\frac1{2n}+\frac1{12n^2}+O\left(\frac1{n^4}\right)\tag{1} $$ Therefore, $$ \begin{align} \sum_{k=n}^{2n}\frac1k &=(\log(2k)+\gamma)-(\log(n-1)+\gamma)+O\left(\frac1n\right)\\ &=\log(2)+O\left(\frac1n\right)\tag{2} \end{align} $$ and $(2)$ shows that $$ \lim_{n\to\infty}\sum_{k=n}^{2n}\frac1k=\log(2)\tag{3} $$

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