# Asymptotic analysis of harmonic series using Calculus

The problem is to proof that Harmonic series $$\sum_{i=1}^n \frac{1}{i} = O(ln \space n)$$

So, I know that $$ln \space n = \int_{1}^n \frac{1}{x} dx$$ so, I need to prove that

$$H(n) = 1+\frac{1}{2}+...+\frac{1}{n} \le c_{1}\int_{1}^n \frac{1}{x} dx$$

Next, I obeserved

$$\frac{1}{2}$$ and $$\int_{1}^2 \frac{1}{x} dx$$

$$\frac{1}{3}$$ and $$\int_{2}^3 \frac{1}{x} dx$$

. . .

$$\frac{1}{n}$$ and $$\int_{n-1}^n \frac{1}{x} dx$$

If I sum all the fraction side, being $$X = \frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}$$, I'll get the equation $$X+1 = H(n)$$

Also, sum of integral side is $$\int_{1}^n \frac{1}{x}dx$$

And I know by using calculator that $$\frac{1}{n}$$ $$\le$$ $$\int_{n-1}^n \frac{1}{x} dx = ln \space n$$

So from the sum of fraction side and integral side, I'll get $$H(n) - 1 \le \int_{1}^n \frac{1}{x}dx$$

$$H(n) \le \int_{1}^n \frac{1}{x}dx +1$$

and by let $$c_1 \ge 2$$, I get my proof that Harmonic series $$\sum_{i=1}^n \frac{1}{i} = O(ln \space n)$$

My problem is I don't understand the mathatical proof saying $$\frac{1}{n}$$ $$\le$$ $$\int_{n-1}^n \frac{1}{x} dx = ln \space n$$

I saw articles of same problems here and know that there are a lots of proof without using calculus. There also the calculus proof (like this) but I don't really understand where to apply it to the equation. (In the link I don't understand about the inequality given)

I want some hint or guide or any resources telling me what to do next to proof this.

It is not true that $$\int_{n-1}^{n} \frac 1 t dt=\ln \, n$$. The correct statement is $$\frac 1 n \leq \int_{1}^{n} \frac 1 t dt =\ln \, n$$. To prove the inequality here observe that $$t \leq n$$ implies $$\frac 1 t \geq \frac 1 n$$. Form this we get $$H(n) \leq \int_{1}^{n} \frac 1 t dt= \ln\, n$$
On the interval between $$n-1$$ and $$n$$, $$\frac1x\ge\frac1n$$ so that $$\int_{n-1}^n\frac{dx}x\ge\int_{n-1}^n\frac{dx}n=\frac1n.$$ But, by calculus, $$\int_{n-1}^n\frac{dx}x=\Big[\ln x\Big]_{x=n-1}^n=\ln n-\ln(n-1)$$ (not $$\ln n$$) since the derivative of $$\ln x$$ is $$1/x$$. But the calculation you really need is $$\int_{1}^n\frac{dx}x=\Big[\ln x\Big]_{x=1}^n=\ln n-\ln1=\ln n.$$