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Let $$A_n:=\frac{1}{n+1}+\cdots+\frac{1}{2n}.$$ It is well known that $A_n$ is an increasing sequence, and $\lim_{n\rightarrow \infty}A_n=\ln(2)$. Motivated by how fast $A_n$ converges to $ln(2)$, I would like to know the limit below $$\lim_{n\rightarrow\infty}n\left(\ln(2)-A_n\right).$$

First, since $1/(1+x)$ is a strictly decreasing function, we have $$\ln 2=\int_0^1 \frac{1}{1+x}\, dx<\frac{1}{n}\sum_{k=0}^{n-1}\frac{1}{1+k/n}=\frac{1}{n}+\cdots+\frac{1}{2n-1}.$$ Therefore, we have $$\ln(2)-A_n<\left(\frac{1}{n}+\cdots+\frac{1}{2n-1}\right)-\left(\frac{1}{n+1}+\cdots+\frac{1}{2n}\right)=\frac{1}{n}-\frac{1}{2n}=\frac{1}{2n}.$$ Hence $B_n:=n(\ln2-A_n)$ is bounded from above, and $$\limsup_{n\rightarrow\infty}B_n\le \frac{1}{2}.$$

Furthermore, since $b_n:=1/(\ln 2-A_n)$ is strictly increasing, by the Stolz theorem, we have $$\lim_{n\rightarrow \infty}B_n=\lim_{n\rightarrow \infty}\frac{1}{b_{n+1}-b_n}=\lim_{n\rightarrow \infty}4(1+\frac{1}{2n})B_nB_{n+1}.$$

Therefore, if $B_n$ has a limit, then the limit has to be $0$ or $1/4$.

However, first, I have some trouble in showing the monotonicity of $B_n$, and more importantly, I don't know how to exclude the possibility that the limit of $B_n$ cannot be $0$.

Any ideas or comments are fully appreciated.

QFL
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    Another way is to use the asymptotic result $\sum_{k = 1}^n {\frac{1}{k}} = \ln n + \gamma + \frac{1}{{2n}} + \mathcal{O}!\left( {\frac{1}{{n^2 }}} \right)$ which gives $n\left( {\ln 2 - \sum_{k = n + 1}^{2n} {\frac{1}{k}} } \right) = \frac{1}{4} + \mathcal{O}!\left( {\frac{1}{n}} \right)$. – Gary Feb 07 '24 at 08:41
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    https://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula – Giulio R Feb 07 '24 at 09:20

4 Answers4

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The result is pretty general:

Lemma: If $f\in C^{1}[0,1] $ then $$\lim_{n\to\infty} \sum_{k=1}^{n}f\left(\frac{k}{n}\right)-n\int_{0}^{1}f(x)\,dx=\frac{f(1)-f(0)}{2}$$ (proof available here)

In the above lemma it is sufficient to assume the Riemann integrability of $f'$.

For your current problem $f(x) =1/(1+x)$ and then the lemma says that $$\lim_{n\to\infty} n\left(\sum_{k=1}^n\frac{1}{n+k}-\log 2\right) =\frac{f(1)-f(0)}{2}$$ so that the desired limit is $1/4$.

Paramanand Singh
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Consider the limit of $$ \lim_{n \to \infty} n \, \left( \ln 2 - \sum_{k=1}^{n} \frac{1}{n+k} \right) = \lim_{n \to \infty} \phi_{n} $$ in the following way. Since $$H_{n} \approx \ln n + \gamma + \frac{1}{2 n} - \frac{1}{12 \, n^2} + \mathcal{O}\left(\frac{1}{n^3}\right) $$ then \begin{align*} \phi_{n} &= n \, \left( \ln 2 - \sum_{k=1}^{n} \frac{1}{n+k} \right) \\ &= n \, \left( \ln 2 - (H_{2n} - H_{n}) \right) \\ &= n \, \left( \ln 2 - \left( \ln 2 - \frac{1}{4 n} + \frac{1}{16 \, n^2} + \mathcal{O}\left(\frac{1}{n^3}\right) \right) \right) \\ &= \frac{1}{4} - \frac{1}{16 \, n} + \mathcal{O}\left(\frac{1}{n^2}\right). \end{align*}

From this is it can be seen that $$ \lim_{n \to \infty} n \, \left( \ln 2 - \sum_{k=1}^{n} \frac{1}{n+k} \right) = \frac{1}{4}. $$

Leucippus
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Another simple way without replying on properties of Harmonic functions (which are not very easy to prove) is to use the Stolz-Cesaro Theorem which is an analogue of the L'Hospital Rule in the discrete case.

Write it as $$\frac{\ln(2)-\sum_{k=1}^{n}\frac{1}{n+k}}{\frac{1}{n}}=\frac{a_{n}}{b_{n}}$$ and we have a $\frac{0}{0}$ form.

You get using the Theorem above that $$\lim_{n\to\infty}\frac{a_{n}-a_{n+1}}{b_{n}-b_{n+1}}=\lim_{n\to\infty}\frac{\sum_{k=1}^{n+1}\frac{1}{n+1+k}-\sum_{k=1}^{n}\frac{1}{n+k}}{\frac{1}{n}-\frac{1}{n+1}}$$

Which is just $$\lim_{n\to\infty}\frac{ \frac{1}{2n+1}+\frac{1}{2n+2}-\frac{1}{n+1}}{\frac{1}{n(n+1)}}=\lim_{n\to\infty}\frac{n(n+1)}{(2n+1)(2n+2)}=\frac{1}{4}$$

Mr.Gandalf Sauron
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  • Thanks for the simple answer. I did not realize there is a version of Stolz theorem for 0/0, but this is indeed correct. – QFL Feb 07 '24 at 09:26
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Notice $\log 2 = \log\frac{2n}{n} = \sum\limits_{m=n}^{2n-1}\log\frac{m+1}{m}$, we have

$$\begin{align} \log 2 - \sum_{k=1}^n\frac{1}{n+k} & = \sum_{m=n}^{2n-1}\int_0^1 \left(\frac{1}{m+x} - \frac{1}{m+1}\right)dx\\ & = \sum_{m=n}^{2n-1}\int_0^1\frac{1-x}{(m+x)(m+1)}dx\\ & = \sum_{m=n}^{2n-1}\int_0^1\left(\frac{1-x}{m(m+1)} - \frac{x(1-x)}{m(m+1)(m+x)}\right)dx\\ &= \sum_{m=n}^{2n-1}\left[\frac12\left(\frac1m - \frac1{m+1}\right) - \int_0^1 \frac{x(1-x)}{m(m+1)(m+x)}dx\right]\\ &= \frac1{4n} - \sum_{m=n}^{2n-1}\int_0^1 \frac{x(1-x)}{m(m+1)(m+x)}dx \end{align} $$ This leads to the bound $$\begin{align} \left|n\left(\log 2 - \sum_{k=1}^n\frac{1}{n+k}\right) - \frac14\right| & = n \sum_{m=n}^{2n-1} \int_0^1 \frac{x(1-x)}{m(m+1)(m+x)}dx\\ & \le \sum_{m=n}^{2n-1} \int_0^1 \frac{x(1-x)}{m(m+1)}dx\\ & = \frac{1}{12n} \end{align} $$ As a result, $$\lim_{n\to\infty} n\left(\log 2 - \sum_{k=1}^n\frac{1}{n+k}\right) = \frac14$$

achille hui
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