Prove $\frac{1}{n} + \frac{1}{n+1} + .... + \frac{1}{2n} > 0.63 $ I need to prove this: 
$\frac{1}{n} + \frac{1}{n+1} + .... + \frac{1}{2n} > 0.63 $.
I know its $\log{(2)} $ and it's something like $\int_{0}^{1} \ \frac{dx}{1+x}\ =  \log{(2)}$, but I don't know the steps between those. 
 A: See: https://en.wikipedia.org/wiki/Harmonic_series_(mathematics)#Integral_test
Similarly you can show that $$\int_{n}^{2n}\frac{1}{x}dx<\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{2n}$$
Notice that: $$\int_{n}^{2n}\frac{1}{x}dx=\log(2n)-\log(n)=\log\left(\frac{2n}{n}\right)=\log(2)$$
A: Let $(a_n)$ be the sequence
$$a_n= \frac{1}{n}+\frac{1}{n+1} + \dots + \frac{1}{2n},$$
then $a_1=1.5$ and $a_{n+1}-a_n= \frac{1}{2n+2} +\frac{1}{2n+1} -\frac{1}{n}<0$, then the sequence is decreasing, as u have noticed that $a_n \to \log  2$, it won't be hard to figure out that $a_n>\log 2 \approx 0.693$ for all integers $n>0$.
A: $$
\begin{align}
\frac1n+\frac1{n+1}+\cdots+\frac1{2n}
&\ge\int_n^{2n+1}\frac1x\,\mathrm{d}x\\
&=\log\left(2+\frac1n\right)\\[3pt]
&\gt\log(2)
\end{align}
$$
A: You can obtain the desired inequality by checking it directly for $n=1$ and $2$ and then proving, by induction, the stronger inequality
$${1\over n}+{1\over n+1}+\cdots+{1\over2n}\gt{2\over3}+{3\over4n-1}\qquad\text{for }n\ge3$$
The base case is
$${1\over3}+{1\over4}+{1\over5}+{1\over6}={19\over20}\gt{31\over33}={2\over3}+{3\over11}$$
(the key calculation is $33\cdot19=627\gt620=20\cdot31$), and the induction step boils down to verifying the inequality
$${3\over4n-1}+{1\over2n+1}+{1\over 2n+2}-{1\over n}\gt{3\over4n+3}$$
which is messy but doable.
In case anyone is curious where the stronger inequality came from, I picked the $2\over3$ as a convenient fraction that's less than $\ln2\approx.69$ and then did a bunch of work to find something of the form $a\over bn+c$ to tack on that would make the induction step work.
