Could someone explain how the Gram series relates to Riemann's function? I was reading an article on the distribution primes which mentions the following equation for Riemann's function $R(x)$:
$$R(x) = \sum_{n=1}^{\infty}\frac{\mu(n)}{n}\text{li}(x^{1/n}) = 1 + \sum_{k=1}^{\infty}\frac{(\ln x)^k}{k k!\zeta(k+1)}$$
Here's what the article says:

The last form above for $R(x)$ is the Graham series and is an
  excellent way to calculate this function.

I checked Wikipedia but I could not find any article on the Gram series.  Could someone provide information on the Gram series, what it is, and how it relates to $R(x)$.

Edit: The article I was reading incorrectly calls the Gram Series, the "Graham Series".  I have changed the spelling in each case except for the quote from the article.
 A: This should be the Gram series but convergence will not be as good as announced since fast for $|\ln x|<1\,$ but slow for $|\ln x|\gg 1$ (with very large terms appearing as $k$ grows as for the exponential expansion). 
Concerning a derivation let's note that $\;\text{li}(x)=\text{Ei}(\ln x)\;$ and (from DLMF $(6.6.1)$
$$\text{Ei}(s)=\gamma+\ln s+\sum_{k=1}^\infty\frac{s^k}{k!\,k}$$
allowing us to go from the first to the second series (this is Berndt's derivation in Ramanujan's Notebooks IV p.$127$):
\begin{align}
R(x) &:= \sum_{n=1}^{\infty}\frac{\mu(n)}{n}\text{li}(x^{1/n})\\
&= \sum_{n=1}^{\infty}\frac{\mu(n)}{n}\text{Ei}\left(\frac 1n\,\ln x\right)\\
&= \sum_{n=1}^{\infty}\frac{\mu(n)}{n}\left(\gamma-\ln n+\ln\ln x+\sum_{k=1}^\infty\frac{(\ln x)^k}{k!k\,n^k}\right)\\
&= (\gamma+\ln\ln x)\sum_{n=1}^{\infty}\frac{\mu(n)}{n}-\sum_{n=1}^{\infty}\frac{\mu(n)}{n}\ln n+\sum_{n=1}^{\infty}\frac{\mu(n)}{n}\sum_{k=1}^\infty\frac{(\ln x)^k}{k!\,k\,n^k},\ (*)\\
&=1+\sum_{k=1}^\infty\frac{(\ln x)^k}{k!\,k}\sum_{n=1}^{\infty}\frac{\mu(n)}{n^{k+1}}\\
& = 1 + \sum_{k=1}^{\infty}\frac{(\ln x)^k}{k k!\zeta(k+1)}\\
\end{align}
$(*)$ Using the well known facts (by Landau) :
\begin{align}
\sum_{n=1}^{\infty}\frac{\mu(n)}{n}&=0\\
\sum_{n=1}^{\infty}\frac{\mu(n)}{n}\ln n&=-1\\
\sum_{n=1}^{\infty}\frac{\mu(n)}{n^s}&=\frac 1{\zeta(s)}\\
\end{align}
It would seem that the two first series could be deduced from the third one (obtained from the Euler product) but the first one is in fact "as deep as" the PNT.
