0%

The harmonic series,
$$1+\frac1{2}+\frac1{3}+\frac1{4}+\frac1{5}+\dots =\sum_{n=1}^\infty\frac1{n}=?$$

Let’s assume that we know $ln’(x)=\frac{d}{dx}ln(x)=\frac1{x}$

Then from above we can see that

$$\sum_{n=0}^\infty >\int_{n=0}^{n=\infty}\frac1{x}dx=ln(n)=\infty$$

So the harmonic series is a divergent series.
In fact, the difference between these two infinite number is a constant (called Euler–Mascheroni constant $\gamma$)

$$\sum_{n=0}^\infty =ln(n)+\gamma$$

But the alternating harmonic series is different,

$$1-\frac1{2}+\frac1{3}-\frac1{4}+\frac1{5}-\dots =-\sum_{n=1}^\infty\frac{(-1)^n}{n}=?$$

There’s a rather counterintuitive way of calculating it.

$$1-\frac1{2}+\frac1{3}-\frac1{4}+\frac1{5}-\dots$$

To generalize the series,

$$\hookrightarrow x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\dots$$

And then take the derivates of it,

\begin{align} \hookrightarrow &\frac{d}{dx}(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\dots)\nonumber\\ &=1-x+x^2-x^3+x^4-\dots\nonumber\\ &=\frac1{1+x}\nonumber \end{align}

And then, to integrate it back, and to specialize it to the series we are interested in,

\begin{align} &1-\frac1{2}+\frac1{3}-\frac1{4}+\frac1{5}-\dots\nonumber\\ &=[x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\dots]_0^1\nonumber\\ &=\int_0^1\frac1{1+x}dx\nonumber\\ &=ln(1+x)|_0^1\nonumber\nonumber\\ &=ln(2)\nonumber \end{align}

Reference:

- YouTube: What makes the natural log “natural”? | Lockdown math ep. 7