Asymptotic formula for $\sum_{n\le x}\frac 1n$

Question

Prove that $\displaystyle \sum_{n\le x}\frac 1n=\log (x+2)+O(1)$.

We have, $\displaystyle \sum_{n\le x}\frac 1n=\log x +\gamma+O(1/x)$. Now if I can show that $\log(x+2)-\log x+O(1/x)=O(1)$ then I have done.

Now, $\log(x+2)-\log x+O(1/x)=\log \left(\frac{x+2}{x}\right)+O(1/x)=O(1/x).$

How can I prove this ?

Answer 1

All you need is $\log(1+z) < z$ for $z > 0$.

Then $\log(x+2)-\log(x) =\log(1+\frac{2}{x}) \lt \frac{2}{x} =O(\frac1{x}) $.

To show $\log(1+z) < z$ for $z > 0$:

$\log(1+z) =\int_0^z \frac{dt}{1+t} \lt \int_0^z dt =z $.

You can also use the integral to get a lower bound:

$\begin{array}\\ \log(1+z) &=\int_0^z \frac{dt}{1+t}\\ &\gt \int_0^z \frac{dt}{1+z}\\ &=\frac{z}{1+z}\\ &=\frac{z+z^2-z^2}{1+z}\\ &=z-\frac{z^2}{1+z}\\ &\gt z-z^2\\ \end{array} $