# Sum of Harmonic Series

This is what I thought about when I was 17 year old. Because when I read about Maclaurin expansion, which is Taylor expansion at zero.

$\displaystyle\log(1+x)=\sum_{k=1}^\infty (-1)^{k-1}\frac{x^k}{k}$,

however, this equality is valid for every $x\in (-1,1]$.

Because of this convergent series, I got to know that:

$\displaystyle\log 2=\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\dots$

Though there is also a elementary way to calculate if we know $\displaystyle\sum_{k=1}^n\frac{1}{k}\sim \log n$.

We shall ask if we need to calculate other series,

$\displaystyle\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{2k-1}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\dots$.

Certainly we can use calculus, it is pretty simple if we treat them as a series function’s calculus’ expansion.

$\displaystyle\int_{0}^{1}\frac{1}{1+x^{p}}$

This technique requires the primary function’s explicit form…it is very difficult.

At that time, I gave my method of substituting $x$ by a complex number, and depart the Re and Im. It really works with low degree $p$.

If I replace $x$ by $\omega$, where $\omega$ satifies $\omega^n=1$. We need prove the validity of the Maclaurin Formula.

We first define the multi-value function $Log(1+z)$, however we can restrict to one branch.

Then we have to know whether its expansion is convergent, because the radius of convergence is 1 actually.

BY ABEL TEST FOR CONVERGENCE,

$\displaystyle a_{k}=\frac{1}{k}$,

$\displaystyle b_{k}=(-1)^{k-1} z^k$.

• $a_{k}$ is monotonic decreasing to zero.
• $\sum_{k=0}^\infty b_{k}$ is convergent whenever $z\neq -1$.

Thus the expansion of $Log(1+z)$ is valid now for $z=\omega$.

Surely we can set $z=i$ or $z=\omega$ which $\omega^3=1$ to see what will happen. First we set $z=i$,

$\displaystyle Log(1+i)=\log\sqrt{2}+\frac{\pi}{4}i$, and

$\displaystyle \sum_{k=1}^\infty \frac{(-1)^{k-1}i^k}{k}=i\sum_{k=1}^\infty \frac{(-1)^k}{2k-1}+\sum_{k=1}^\infty \frac{(-1)^{k-1}}{2k}$.

Then it is done.

If we set it to be cubic root of 1, it is more complicated, but still calculable.

The process we shall omit, and we give the answer directly.

$\displaystyle Log(1+\omega)=\sum_{k=1}^{\infty}(-1)^{k+1}\frac{\omega}{3k-2}+\sum_{k=1}^{\infty}(-1)^{k+1}\frac{\omega^2}{3k-1}+\sum_{k=1}^\infty (-1)^{k+1}\frac{1}{3k}=\omega I+\omega^2J+K$.

$\omega=\displaystyle -\frac{1}{2}+\frac{\sqrt{3}}{2}i$, thus the formula turns into:

$\displaystyle Log(1+\omega)=-\frac{1}{2}I-\frac{1}{2}J+K+\frac{\sqrt{3}}{2}i(I-J)$

And $I+J+K=\log 2$,

$\displaystyle Log(1+\omega)=\frac{\pi}{3}i$

Therefore,

$\displaystyle I-J=\frac{2\sqrt{3}}{9}\pi$,

$\displaystyle -\frac{1}{2}I-\frac{1}{2}J+K=0$,

$I+J+K=\log 2$.

Solve it and $I=\displaystyle \frac{1}{3}\log 2+\frac{\sqrt{3}}{9}\pi$, $J=\displaystyle \frac{1}{3}\log 2-\frac{\sqrt{3}}{9}\pi$, $\displaystyle K=\frac{1}{3}\log 2$.

It is done. However, for higher degree, we could not always find a explicit answer to it (due to Guass’ proof), but if the root $\omega$ can be explicitly written, then we can solve it.

$\Box$