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Is it the case that

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

That is, does the "second half" of the harmonic series go to zero?

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$$\sum_{n=k+1}^{2k}\frac1n\ge\sum_{n=k+1}^{2k}\frac1{2k}=\frac12,$$ so it obviously won't tend to zero. Have you heard of Riemann sums? – Jyrki Lahtonen Oct 18 '11 at 4:07
Can we agree that this sum is at least $\sum_{n=k+1}^{2k} \frac{1}{2k} = \frac{1}{2}$ and at most $\sum_{n=k+1}^{2k} \frac{1}{k} = 1$? – Qiaochu Yuan Jun 7 '12 at 18:22
@QiaochuYuan: what sum? – T. Eskin Jun 7 '12 at 18:35
@Thomas: $\sum_{n=k+1}^{2k} \frac{1}{n}$. The above bounds apply for every value of $k$. – Qiaochu Yuan Jun 7 '12 at 18:36
Before finding a rigorous answer, there is a way to obtain $\ln(2)$ as a conjectured value. We know the dominating term of the asymptotic growth of the harmonic series is $\ln(n)$. For large enough values of $n$, the difference between the harmonic series up to $2n$ and $n$ is $\ln(2n) - \ln(n) = \ln(2) + \ln(n) - \ln(n) = \ln(2)$. ;) – CosmoVibe Jul 23 at 0:58

8 Answers 8

up vote 38 down vote accepted

The summation you have written converges to $\log(2)$.$$\lim_{k \rightarrow \infty} \sum_{n=k+1}^{2k} \frac1n = \lim_{k \rightarrow \infty} \left( \sum_{n=1}^{2k} \frac1n - \sum_{n=1}^{k} \frac1n\right) = \lim_{k \rightarrow \infty} \left( \sum_{n=1}^{2k} \frac1n - \log(2k) - \sum_{n=1}^{k} \frac1n + \log(k) + \log(2) \right).$$ Note that $$\lim_{k \rightarrow \infty } \left(\sum_{n=1}^{k} \frac1n - \log(k) \right) = \gamma.$$ Let $\displaystyle a_k = \left(\sum_{n=1}^{k} \frac1n - \log(k) \right)$ and we have $\displaystyle \lim_{k \rightarrow \infty} a_k = \gamma$. Hence, the summation you have can be written as $$\lim_{k \rightarrow \infty} \sum_{n=k+1}^{2k} \frac1n = \lim_{k \rightarrow \infty} \left(a_{2k} -a_k + \log(2) \right) = \gamma - \gamma + \log(2) = \log(2)$$

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Hint: $\sum_{n=1}^N \frac{1}{n} = \ln(N) + \gamma + O(1/N)$

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Well, no, the limit is $\log 2.$ The basic fact is that the finite sum $$ \sum_{m = 1}^W \frac{1}{m} \approx \gamma + \log W,$$ where $\gamma \approx 0.5772156649\ldots$ is the Euler-Mascheroni constant. So take the approximation for $W= 2 k$ and subtract the approximation for $W=k.$

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Rewriting the sum as $$ \sum_{n=k+1}^{2k}\frac1n=\sum_{n=k+1}^{2k}\frac1k\cdot\frac1{n/k} $$ allows us to identify this as a Riemann sum related to the definite integral $$\int_1^2\frac1x\,dx=\ln 2.$$ To see that, divide the interval $[1,2]$ to $k$ equal length subintervals, and evaluate the function $f(x)=1/x$ at the right end of each subinterval. When $k\to\infty$, the Riemann sums will then tend to the value of this definite integral.

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This is nice Jyrki. I would not have thought about it in this way! – anthus Oct 18 '11 at 4:46
Thanks @Daniel. My motivation to post this as an answer was that the earlier answers all resorted to asymptotics and Euler-Mascheroni constant, which felt like too high-powered a tool. – Jyrki Lahtonen Oct 18 '11 at 5:14
Good intuition, +1. – Oo3 Jun 7 '12 at 19:35
This has been getting sporadic upvotes since the election. Thank you for your support, but I'm switching this to CW. – Jyrki Lahtonen Jan 2 at 15:24

The sum is equal to $A_n = (1/1 - 1/2 + 1/3 \dots -1/2n)$.

As an alternating series, it satisfies $|A_n - \log 2| < \frac{1}{2n}$.

The asymptotics of harmonic numbers, using Euler's constant, are not needed to get the $O(1/n)$ convergence or its extension to higher powers of $1/n$.

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Starting from A_n may be simpler to see. $$A_n = H_{2n} - 2(1/2 + 1/4 + \dots +1/2n) = H_{2n} - H_n$$ which is the sum of interest. The sign is corrected, thanks (@Jyrki). – zyx Oct 18 '11 at 5:32
Oops. I missed that way of looking at it :-) – Jyrki Lahtonen Oct 18 '11 at 5:37

More generally, for $p \geq q>1$ one has

$$\lim\limits_{n \to \infty} \sum\limits_{k=qn+1}^{np} \frac{1}{k}=\log \frac{p}{q}$$

which can be proven using

$$\lim\limits_{n \to \infty} \sum\limits_{k=1}^{np} \frac{1}{k}-\log (pn)=\gamma$$

$$\lim\limits_{n \to \infty} \sum\limits_{k=1}^{nq} \frac{1}{k}-\log (qn)=\gamma$$

in the same spirit as Sivaram's answer.

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We may recall the celebre limit that yields Euler-Mascheroni constant, namely:

$$\lim_{n\to\infty} 1+\frac1{2}+\cdots+\frac{1}{n}-\ln{n}={\gamma}$$ $\tag{$\gamma$ is Euler-Mascheroni constant}$ Then everything boils down to: $$\lim_{n\to\infty}\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n} = \lim_{n\to\infty}{\gamma}+\ln{2n}-{\gamma}-\ln{n}= \ln{2}.$$


Use one of the consequences of the Lagrange's theorem applied on $\ln(x)$ function, namely:

$$\frac{1}{k+1} < \ln(k+1)-\ln(k)<\frac{1}{k} \space , \space k\in\mathbb{N} ,\space k>0$$

Taking $k=n,n+1,...,2n$ values to the inequality and then summing all relations, we get all we need in order to apply Squeeze theorem.


We may use Botez-Catalan identity and immediately get that:

$$\lim_{n\to\infty}\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n} = \lim_{n\to\infty} 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + (-1)^{2n+1}\frac{1}{2n}= $$ $$\lim_{n\to\infty} 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + (-1)^{n+1}\frac{1}{n}=\ln{2}.$$ The last series' limit is obtained by using Taylor expansion of $\ln(x+1)$ and take $x=1$

The proofs are complete.

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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ Is $\ds{\lim_{k \to \infty}\sum_{n = k + 1}^{2k}{1 \over n} = 0\ {\large ?}}$

\begin{align} \color{#c00000}{\lim_{k \to \infty}\sum_{n = k + 1}^{2k}{1 \over n}}&= \lim_{k \to \infty}\sum_{n = 0}^{k - 1}{1 \over n + k + 1} =\lim_{k \to \infty}\sum_{n = 0}^{k - 1}\int_{0}^{1}t^{n + k}\,\dd t =\lim_{k \to \infty}\int_{0}^{1}\sum_{n = 0}^{k - 1}t^{n + k}\,\dd t \\[3mm]&=\lim_{k \to \infty}\int_{0}^{1}{t^{k}\pars{t^{k} - 1} \over t - 1}\,\dd t =-\lim_{k \to \infty}\int_{0}^{1} \ln\pars{1 - t}\bracks{2kt^{2k - 1} - kt^{k - 1}}\,\dd t \end{align}

$$ \color{#c00000}{\lim_{k \to \infty}\sum_{n = k + 1}^{2k}{1 \over n}} =\lim_{k \to \infty}\bracks{{\cal F}\pars{k} - {\cal F}\pars{2k}} \quad\mbox{where}\quad {\cal F}\pars{k}\equiv k\int_{0}^{1}\ln\pars{1 - t}t^{k - 1}\,\dd t\tag{1} $$

Let's evaluate ${\cal F}\pars{k}$: \begin{align} \color{#c00000}{{\cal F}\pars{k}}&=k\lim_{\mu \to 0}\partiald{}{\mu} \int_{0}^{1}\pars{1 - t}^{\mu}t^{k - 1}\,\dd t =k\lim_{\mu \to 0}\partiald{}{\mu}\bracks{% \Gamma\pars{\mu + 1}\Gamma\pars{k} \over \Gamma\pars{\mu + 1 + k}} \\[3mm]&=\Gamma\pars{k + 1}\braces{{\Gamma\pars{1} \over \Gamma\pars{1 + k}} \bracks{\Psi\pars{1} - \Psi\pars{1 + k}}}=\color{#c00000}{% \Psi\pars{1} - \Psi\pars{1 + k}} \end{align}

With expression $\pars{1}$: \begin{align} \color{#00f}{\large\lim_{k \to \infty}\sum_{n = k + 1}^{2k}{1 \over n}} &=\lim_{k \to \infty}\bracks{\Psi\pars{2k + 1} - \Psi\pars{k + 1}} =\lim_{k \to \infty}\bracks{\ln\pars{2k + 1} - \ln\pars{k + 1}} \\[3mm]&=\lim_{k \to \infty}\ln\pars{2k + 1 \over k + 1} =\color{#00f}{\large\ln\pars{2}} \not=0 \end{align}

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