How prove this $H_{2n}-H_{n}+\frac{1}{4n}>\ln{2}$

Question

Show that, for every positive integer $n$, $$\dfrac{1}{n+1}+\dfrac{1}{n+2}+\cdots+\dfrac{1}{2n}+\dfrac{1}{4n}>\ln{2}$$

I know this $$\lim_{n\to\infty}\dfrac{1}{n+1}+\dfrac{1}{n+2}+\cdots+\dfrac{1}{2n}=\ln2$$

and use this $$\ln{(1+\dfrac{1}{n})}<\dfrac{1}{n}$$ is not useful

But for this inequality I can't. Thank you

Answer 1

Let $A_n=H_{2n}-H_{n}+\frac{1}{4n}$, then a tedious but easy computation shows that $$ A_{n+1}-A_n=\frac{-1}{4n(n+1)(2n+1)}\lt0, $$ hence $(A_n)$ is decreasing, in particular, $A_n\gt A$ where $A=\lim\limits_{n\to\infty}A_n$. Since $H_n=\log n+\gamma+o(1)$ when $n\to\infty$, $A=\log2$. QED.

Remark: The funny thing here (and the subject of the next questions of the exercise as it is usually asked) is that the same technique applies to the sequence $B_n=H_{2n}-H_{n}$, showing that $(B_n)$ is increasing, hence $B_n\lt B$ where $B=\lim\limits_{n\to\infty}B_n$. Since $B=\log2$ as well, this yields some approximations of $\log2$, namely the fact that, for every $n\geqslant1$, $$ H_{2n}-H_{n}\lt\log2\lt H_{2n}-H_{n}+\frac1{4n}. $$ Likewise, for every $n\geqslant1$, $$ H_{2n}-H_{n}+\frac1{4n+2}\lt\log2\lt H_{2n}-H_{n}+\frac1{4n}, $$ thus, the correction $\frac1{4n}$ is asymptotically of the right order.

Answer 2

Write $$u_n = \frac{1}{n+1} + \frac{1}{n+2} + \dots + \frac{1}{2n} + \frac{1}{4n}$$

$$ = [\frac{1}{1} + \frac{1}{2} + \dots+ \frac{1}{4n}] - [\frac{1}{1} + \frac{1}{2} + \dots+ \frac{1}{n}] - [\frac{1}{2n+1} + \frac{1}{2n+2} + \dots +\frac{1}{4n -1}]$$

$$= [\gamma_{4n} + \log(4n)] - [\gamma_n + \log(n)] - [\gamma_{4n -1} + \log(4n-1) - \gamma_{2n} - \log(2n)]$$

Now take $\lim_{n \rightarrow \infty} u_n$.

All $\gamma_{n}$, $\gamma_{4n - 1}$, $\gamma_{2n}$, $\gamma_{n}$ etc will be $\gamma$ (Euler's constant) and will be cancelled out.

What remains?

$\lim_{n \rightarrow \infty}[\log{\frac{4n}{4n - 1}} + \log{\frac{2n}{n}}] = \log 2$

Now see $u_{n+1} - u_n = \frac{1}{4(n+1)} - \frac{1}{4n} + \frac{1}{2(n+1)} - \frac{1}{n+1} < 0$

Thus the sequence $\{u_n\}$ is monotone decreasing and converges to $\log 2$. So for all $n$, $u_n > \log 2$

Thus ultimately we are getting for any $n$

$$\frac{1}{n+1} + \frac{1}{n+2} + \dots + \frac{1}{2n} + \frac{1}{4n} > \log 2$$

Answer 3

A not at all tedious computation shows that the expression - which I rearrange a tiny bit -

$$\begin{align} \frac{1}{2n} + \frac{1}{n+1} + \dotsc + \frac{1}{2n-1} + \frac{1}{4n} &= \sum_{k=0}^{n-1} \frac{1}{2}\left(\frac{1}{n+k} + \frac{1}{n+k+1}\right)\\ &= \frac{1}{n} \sum_{k=0}^{n-1} \frac{1}{2}\left(\frac{1}{1 + \frac{k}{n}} + \frac{1}{1+\frac{k+1}{n}}\right) \end{align}$$

is a trapezium sum for the integral

$$\int_0^1 \frac{dt}{1+t} = \log 2,$$

and since $\frac{1}{1+t}$ is convex, the trapeziums have greater area than the corresponding part of the integral.