How can I see that $\sum_2^n \frac{1}{j} = \int_1^n \frac{1}{j}\ dj$?

Question

Can your show how this comes about or give me pointers of investigation so can I work it out? Thank you!

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EDIT1: Various pointed out the equation is false, but that it can be taken as approximation. Since no answered were added, I'd like to ask how I could start from the left hand side and construct an approximation to the sum by using the right hand side. What steps should I take to see how to do that?

Answer 1

More tenuous, you can say $$ \begin{align} \sum_{j=1}^n\dfrac{1}{j} &=\ln n+\gamma+\varepsilon_n\\ \sum_{j=2}^n\dfrac{1}{j} &=\int_1^n\dfrac{1}{j}dj\color{red}{+\gamma+\varepsilon_n}-1 \end{align} $$ where $\varepsilon_n\sim\dfrac{1}{2n}$. You may see the details here.

Answer 2

$\int_{j=1}^n\frac{1}{x}dx=\sum_{j=2}^n\int_{j-1}^j\frac{1}{x}dx$. However $\int_{j-1}^j\frac{1}{x}dx \approx \frac{j-\frac{1}{2}}{j(j-1)}\approx \frac{1}{j}$, giving an approximation to the result you want.

The last step could be made a little clearer. $\frac{1}{j-1}\gt \int_{j-1}^j\frac{1}{x}dx\gt \frac{1}{j}$. Therefore $\int_{j-1}^j\frac{1}{x}dx\approx \frac{1}{j}$.