What is the sum of the 'second half' of the harmonic series?

$$\lim_{k\to\infty}\sum\limits_{n=k+1}^{2k}{\frac{1}{n}} =~ ?$$

More precisely, what is the limit of the above sequence of partial sums?

  • 12
    $\begingroup$ $$\sum_{n=k+1}^{2k}\frac1n\ge\sum_{n=k+1}^{2k}\frac1{2k}=\frac12,$$ so it obviously won't tend to zero. Have you heard of Riemann sums? $\endgroup$ Commented Oct 18, 2011 at 4:07
  • 22
    $\begingroup$ Can we agree that this sum is at least $\sum_{n=k+1}^{2k} \frac{1}{2k} = \frac{1}{2}$ and at most $\sum_{n=k+1}^{2k} \frac{1}{k} = 1$? $\endgroup$ Commented Jun 7, 2012 at 18:22
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    $\begingroup$ @Thomas: $\sum_{n=k+1}^{2k} \frac{1}{n}$. The above bounds apply for every value of $k$. $\endgroup$ Commented Jun 7, 2012 at 18:36
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    $\begingroup$ 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 $\ln(2n) - \ln(n) = \ln(2) + \ln(n) - \ln(n) = \ln(2)$. ;) $\endgroup$
    – zhuli
    Commented Jul 23, 2015 at 0:58
  • 2
    $\begingroup$ I edited the title to make the question easier to find, since it's become the main reference for this frequently asked question. Please consider moving the checkmark to the top-voted answer, which is really the canonical answer to the question (some students encounter the above limit well before they know what $\gamma$ is, and in any case the existence of $\gamma$ is not needed). $\endgroup$
    – user147263
    Commented Jan 9, 2016 at 17:43

12 Answers 12


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.

  • 8
    $\begingroup$ This is nice Jyrki. I would not have thought about it in this way! $\endgroup$ Commented Oct 18, 2011 at 4:46
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    $\begingroup$ 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. $\endgroup$ Commented Oct 18, 2011 at 5:14
  • 2
    $\begingroup$ This has been getting sporadic upvotes since the election. Thank you for your support, but I'm switching this to CW. $\endgroup$ Commented Jan 2, 2015 at 15:24
  • $\begingroup$ Why $\frac{1}{n/k}$ is equal to $dx$ ? $\endgroup$
    – ZchGarinch
    Commented Sep 8, 2023 at 9:06
  • 1
    $\begingroup$ @ZchGarinch With $f(x)=1/x$ we have $f(n/k)=1/(n/k)$. The length of the subinterval is $1/k$ and you can call that $dx$, if you want to. $\endgroup$ Commented Sep 8, 2023 at 10:18

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 $$\lim_{k \rightarrow \infty} \sum_{n=k+1}^{2k} \frac1n = \lim_{k \rightarrow \infty} \left(a_{2k} -a_k + \log(2) \right) = \gamma - \gamma + \log(2) = \log(2)$$



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.


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 $W=k.$


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 $1/n$.

  • 7
    $\begingroup$ 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). $\endgroup$
    – zyx
    Commented Oct 18, 2011 at 5:32
  • $\begingroup$ Oops. I missed that way of looking at it :-) $\endgroup$ Commented Oct 18, 2011 at 5:37

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


I thought it might be instructive to present three alternative approaches that seem to have been omitted. To that end, we proceed.

METHODOLOGY $1$: Integral Bounds and the Squeeze Theorem

Since $f(x)=\frac1x$ is monotonically decreasing on $[k+1,2k]$, then

$$\log\left(\frac{2k+1}{k+1}\right)=\int_{k+1}^{2k+1}\frac1x\,dx\le \sum_{n=k+1}^{2k}\frac1n\le \int_{k}^{2k}\frac 1x\,dx=\log\left(2\right)$$

whereupon application of the squeeze theorem reveals

$$\bbox[5px,border:2px solid #C0A000]{\lim_{k\to \infty}\sum_{n=k+1}^{2k}\frac1n=\log(2)}$$

METHODOLOGY $2$: Using the Taylor Series for $\log(1+x)$ at $x=1$

Note that we can write

$$\begin{align} \sum_{n=k+1}^{2k}\frac1n&=\color{blue}{\sum_{n=1}^{2k}\frac1n}-\color{red}{\sum_{n=1}^k\frac1n}\\\\ &=\color{blue}{\sum_{n=1}^k\left(\frac{1}{2n-1}+\frac{1}{2n}\right)}-\color{red}{2\sum_{n=1}^k\frac{1}{2n}}\\\\ &=\sum_{n=1}^k\left(\frac{1}{2n-1}-\frac1{2n}\right)\\\\ &=\sum_{n=1}^{2k}\frac{(-1)^{n-1}}{n} \tag 1 \end{align}$$

Comparing the Taylor series $\displaystyle\log(1+x)=\lim_{k\to \infty}\sum_{n=1}^k \frac{(-1)^{n-1}x^n}{n}$ at $x=1$ to the right-hand side of $(1)$, we see immediately that

$$\bbox[5px,border:2px solid #C0A000]{\lim_{k\to \infty}\sum_{n=k+1}^{2k}\frac1n=\log(2)}$$

METHODOLOGY $3$: Apply the Euler-Maclaurin Summation Formula

Using the Euler-Maclaurin Summation Formula, we can write

$$\begin{align} \sum_{n=k+1}^{2k}\frac1n&=\int_k^{2k}\frac1x\,dx+\frac12\left(\frac{1}{2k}-\frac{1}{k+1}\right)+O\left(\frac1{k^2}\right)\\\\ &=\log(2)+O\left(\frac1k\right) \end{align}$$

from which we obtain the coveted limit

$$\bbox[5px,border:2px solid #C0A000]{\lim_{k\to \infty}\sum_{n=k+1}^{2k}\frac1n=\log(2)}$$

as expected!

  • $\begingroup$ In M. 1, you forgot the $\frac{1}{x}\,dx$ in the integral for the upper bound. In my opinion, it is nicer if you take $\int_k^{2k} \frac{dx}{x} = \log 2$ as the upper bound there. Your M. 2 is a more detailed version of zyx's answer, which you seem to have overlooked (but don't take that as a reason to remove it). $\endgroup$ Commented Oct 13, 2016 at 13:01
  • $\begingroup$ @danielfischer Daniel, much appreciative of the comment. I've edited accordingly. -Mark $\endgroup$
    – Mark Viola
    Commented Oct 13, 2016 at 13:44

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.

  • $\begingroup$ As a series, instead of a limit: $$ \log\left(\frac{p}{q}\right)=\sum_{i=0}^\infty \left(\sum_{j=pi+1}^{p(i+1)}\frac{1}{j}-\sum_{k=qi+1}^{q(i+1)}\frac{1}{k}\right) $$ which expresses Riemann rearrangements of the (conditionally convergent) cancelling harmonic series $$0=1-1+\frac{1}{2}-\frac{1}{2} +\frac{1}{3}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+...$$ math.stackexchange.com/a/1602987/134791 $$$$ taking $p$ positive terms and $q$ negative terms at each step. $\endgroup$ Commented Feb 1, 2016 at 20:54

For every $x\in [n,n+1]$, with $n\ge1$, we have $\displaystyle \frac{1}{n+1} \le \frac{1}{x} \le \frac{1}{n}$. If we integrate over $[n,n+1]$, we get \begin{align} \int_{n}^{n+1}\frac{1}{n+1}dx &\le& \int_{n}^{n+1}\frac{1}{x}dx &\le& \int_{n}^{n+1}\frac{1}{n}dx\\ \frac{x}{n+1}\Big|_{n}^{n+1} &\le& \ln(x)\Big|_{n}^{n+1}&\le& \frac{x}{n}\Big|_{n}^{n+1}\\ \frac{n+1-n}{n+1} &\le& \ln(n+1)-\ln(n)&\le& \frac{n+1-n}{n}\\ \end{align} that is $$ \frac{1}{n+1} \le \ln(n+1)-\ln(n) \le \frac{1}{n} $$ Taking the sum from $k+1$ to $2k$, we get $$ S_{k}+\frac{1}{2k+1}-\frac{1}{k+1} \le \ln\left(\frac{2k+1}{k+1}\right) \le S_{k}, $$ where $$ S_{k}=\sum_{n=k+1}^{2k}\frac{1}{n} $$ After rearranging, we get $$ \ln\left(\frac{2k+1}{k+1}\right) \le S_{k} \le \ln\left(\frac{2k+1}{k+1}\right)+\frac{1}{k+1}-\frac{1}{2k+1}. $$ Taking the limit, and using the Squeeze Theorem, we deduce that $$ \lim_{k\to \infty}S_{k}=\ln(2). $$

  • $\begingroup$ I think this is the best answer because it is relatively elementary, self-contained, and only needs $\ln(x)=\int_1^x dt/t$. $\endgroup$ Commented May 7, 2018 at 19:18

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

\begin{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 \\[3mm] & {\cal F}\pars{k}\equiv k\int_{0}^{1}\ln\pars{1 - t}t^{k - 1}\,\dd t\tag{1} \end{align}

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}}} \\[3mm] & =\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}} \\[3mm] & =\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}


Using harmonic numbers $$S_k=\sum_{n=k+1}^{2k}{\frac{1}{n}}=H_{2 k}-H_k$$ Now, using, that for large values of $m$ $$H_m=\gamma +\log(m) +\frac{1}{2 m}-\frac{1}{12 m^2}+O\left(\frac{1}{m^4}\right)$$ we get $$S_k=\log (2)-\frac{1}{4 k}+\frac{1}{16 k^2}+O\left(\frac{1}{k^4}\right)$$ which shows the limit and how it is approached.

For illustration purposes, let us use $k=10$; the exact value is $$S_{10}=\frac{155685007}{232792560}\approx 0.6687714$$ while the above approximation gives $$S_{10}\approx \log (2)-\frac{39}{1600}\approx 0.6687722$$

The next term of the expansion being $-\frac{1}{128 k^4}$ this gives than an error of $-\frac{1}{1280000}\approx -7.812500\times 10^{-7}$ as observed in the above example.


Here is still another method to calculate the limit

$$s=\lim_{k\to \infty } \, \left(\sum _{n=k+1}^{2 k} \frac{1}{n}\right)=\log (2)$$

First of all, from the definitions, we identify the partial sum as a difference of harmonic numbers

$$s_k=\sum _{n=k+1}^{2 k} \frac{1}{n}= H_{2k}-H_k$$

Now we form the "generating" sum of the partial sums

$$g(x) = \sum_{k=1}^\infty s_k x^k$$

Using the formula for the generating function of the harmonic number

$$h_1(x) = \sum_{k=1}^\infty H_k x^k = -\frac{\log(1-x)}{1-x}$$


$$h_2(x) = \sum_{k=1}^\infty H_{2k} x^k=\frac{1}{2} \left(\frac{\log \left(1-\sqrt{x}\right)}{\sqrt{x}-1}-\frac{\log \left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right)$$

we find

$$g(x) = h_1+h_2=\frac{1}{2} \left(\frac{\log \left(1-\sqrt{x}\right)}{\sqrt{x}-1}-\frac{\log \left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right)+\frac{\log (1-x)}{1-x}$$

Assuming that the limit $s = c$ exists we would have

$$g(x) = \sum_{k=1}^\infty c x^k = \frac{c}{1-x}$$

Hence our limit is

$$s = \lim_{x \to 1 }g(x)(1-x) \, $$

which is easily calculated to be $\log(2)$.


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