Logarithmic Integral as Non-Convergent Series

Theorem
The Integral Logarithm (i.e. Logarithmic Integral) can be defined in terms of a non-convergent serie. That is:
 * $\displaystyle {\rm li}\ z = \sum_{i\mathop = 0}^{+\infty} \frac{i!\,z}{\ln^{i+1} z}

= \frac{z}{\ln z} \left(\sum_{i\mathop = 0}^{+\infty}\frac{i!}{\ln^i z}\right)$

Proof
From the definition of the integral logarithm, we have:
 * $\displaystyle {\rm li}\ z = \int_0^z \frac{\mathrm dt}{\ln t}$

Using Integration by Parts:

We can continue as long as we want, so we will analyze the comportment of the part out of the integral, and the part in the integral after a number $n$ of iterations of integration by parts (on the same way as we saw previously).

Let $u_n$ be the part out of the integral, and $v_n$ be the part in the integral. After $n$ iterations of the integration by parts theorem, we have:
 * $\displaystyle {\rm li}\ z = u_n + \int_0^z v_n\,\mathrm dt$
 * $\displaystyle u_0 = 0$
 * $\displaystyle v_0 = \frac{1}{\ln t}$

It follows that:
 * $\displaystyle {\rm li}\ z = u_n + \left[t\,v_n\right]_0^z - \int_0^z t\,\mathrm d\left(v_n\right)

= u_n + \left[t\,v_n\right]_0^z + \int_0^z -t\,\mathrm d\left(v_n\right)$ Which gives us the recurrence relations:


 * By recurrence on $n$, with the following recurrence hypothesis:


 * When $n = 0$, we have:
 * $\displaystyle v_0 = \frac{1}{\ln t} = \frac{0!}{\ln^{0+1} t}$
 * Which verifies the hypothesis.
 * By supposing true at $n$, we have at $n+1$:


 * So $\left({\rm R.H.}\right)$ is verified at $n+1$ if it is verified at $n$, so it is proved for every $n\in\N$ (since it is true at $n=0$):


 * By taking $\left({\rm 1}\right)$, and inserting $\left({3}\right)$ in, a new expression for $u_{n+1}$ in function of $u_n$ (recursive expression):
 * $\displaystyle u_{n+1} = u_n + \left[{t\cdot\frac{n!}{\ln^{n+1} t}}\right]_0^z$
 * $\displaystyle u_{n+1} = u_n + \frac{z\,n!}{\ln^{n+1} z} - \frac{0\cdot n!}{\ln^{n+1} 0}

= u_n + \frac{z\,n!}{\ln^{n+1} z}$
 * That is, we can write by expanding: