# Showing that $\sum_{i=1}^n \frac{1}{i} \geq \log{n}$

I have been trying to prove this by induction on $n\in \mathbb{N}$, but this approach seemed to get me nowhere. I have a suspicion it might be necessary to express $\log{n}$ as $\int_1^n 1/x\text{ }dx$, but I could not build a rigorous argument from that.

Any help greatly appreciated!

$$e^{H_n}=e^{1/1}e^{1/2}\cdots e^{1/n}\color{Red}{\gt}\left(1+\frac 11\right)\left(1+\frac 12\right)\cdots\left(1+\frac 1n\right)=n+1\gt n$$ $$\color{Red}{e^x\gt1+x}\tag{x\gt0}$$

• Why it is equal to $n +1$? – user426277 Jun 6 '17 at 12:34

Hint: The sum on the right is equivalent to an LRAM approximation with $\Delta x = 1$ of $\int_1^n \frac{1}{x} dx$. Since $\frac{1}{x}$ is decreasing, LRAM over-estimates.

If you didn't know, LRAM = Left Rectangle Approximation Method. To estimate the area of a function $f(x)$ from $1$ to $n$ with $\Delta x = c$, we find $c(f(1) + f(1+c) + f(1+2c) + \dots + f(n-c))$

Here's a handy visual, taken from the wikipedia page for the Euler-Mascheroni constant

(How do I incorporate this directly in my post?)

• You may want to state what LRAM is short for for OP's sake. – Cameron Williams Mar 6 '14 at 0:16
• I updated it, along with a handy dandy visual. – MCT Mar 6 '14 at 0:24
• Thanks! I have got this picture in mind, but I have trouble turning it into a rigorous argument. – Vitaly B Mar 6 '14 at 0:31