The limit of truncated sums of harmonic series, $\lim\limits_{k\to\infty}\sum_{n=k+1}^{2k}{\frac{1}{n}}$ 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?
 A: 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
$$
\lim_{k\to \infty}S_{k}=\ln(2).
$$
A: 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 $$\lim_{k \rightarrow \infty} \sum_{n=k+1}^{2k} \frac1n = \lim_{k \rightarrow \infty} \left(a_{2k} -a_k + \log(2) \right) = \gamma - \gamma + \log(2) = \log(2)$$
A: $\newcommand{\+}{^{\dagger}}
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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}
A: 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.
A: METHOD I
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: 
$$\lim_{n\to\infty}\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n} = \lim_{n\to\infty}{\gamma}+\ln{2n}-{\gamma}-\ln{n}= \ln{2}.$$
METHOD II
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.
METHOD III
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.
A: 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.
A: 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 $W=k.$ 
A: 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 $1/n$.
A: Hint: $\sum_{n=1}^N \frac{1}{n} = \ln(N) + \gamma + O(1/N)$
A: 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.
A: 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!
A: 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
$$h_1(x) = \sum_{k=1}^\infty H_k x^k = -\frac{\log(1-x)}{1-x}$$
and
$$h_2(x) = \sum_{k=1}^\infty H_{2k} x^k=\frac{1}{2} \left(\frac{\log \left(1-\sqrt{x}\right)}{\sqrt{x}-1}-\frac{\log \left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right)$$
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 $\log(2)$.
