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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.

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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 $H(n) \leq \int_{1}^{n} \frac 1 t dt= \ln\, n$

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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 $$\int_{1}^n\frac{dx}x=\Big[\ln x\Big]_{x=1}^n=\ln n-\ln1=\ln n.$$

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