<s>[INST] 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?
 [/INST] 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
$$
\lim_{k\to \infty}S_{k}=\ln(2).
$$
</s>