Formula for Natural logarithm of $\pi$ Does any formula or expansion exist that gives $\ln \pi$ ?
The expansion should not just be any formula of $\pi$ with a $\ln$ before it. For example $\ln \pi$ = k + $\sum f(x)$ or something of this type.

Let the matrix $A$ be the lower triangular matrix defined as:
$$A=\text{If } n \bmod k=0 \text{ then } \frac{1}{n^c} \text{ else } 0$$
and let the matrix $B$ be the upper triangular matrix defined as:
$$B=\text{If } k \bmod n=0 \text{ then } \frac{n \cdot \mu (n)}{k^s} \text{ else } 0$$
where $\mu(n)$ is the Möbius function.
Then the matrix product of the matrices $A$ and $B$ is the symmetric matrix $T$ starting:
$$T=A.B = \left(   \begin{array}{ccccccc}   +1&+1&+1&+1&+1&+1&+1&\cdots \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&-2&+1&+1&-2&+1 \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&+1&+1&-4&+1&+1 \\ +1&-1&-2&-1&+1&+2&+1 \\ +1&+1&+1&+1&+1&+1&-6 \\ \vdots&&&&&&&\ddots \end{array}   \right)$$
which by analytic continuation has the Dirichlet generating function:
$$\sum\limits_{k=1}^{\infty}\sum\limits_{n=1}^{\infty} \frac{T(n,k)}{n^c \cdot k^s} = \sum\limits_{n=1}^{\infty} \frac{\lim\limits_{z \rightarrow s} \zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(z-1)}}}{n^c} = \frac{\zeta(s) \cdot \zeta(c)}{\zeta(c + s - 1)}$$
Then take the limit:
$$\frac{\zeta '(s)}{\zeta (s)}=\lim_{c\to 1} \, \left(\zeta (c)-\frac{\zeta (c) \zeta (s)}{\zeta (c+s-1)}\right)$$
which is equivalent to subtracting the first column in the matrix $T$.
Evaluate at $s=0$
$$\frac{\zeta '(0)}{\zeta (0)}=\log(2\pi)$$
subtract with $\log(2)$
$$\frac{\zeta '(0)}{\zeta (0)}-\log(2) = \log(\pi)$$
So by analytic continuation:
$$\log(\pi)=\sum\limits_{k=1}^{\infty}\sum\limits_{n=1}^{\infty} \frac{T_1(n,k)}{n^1 \cdot k^0}-\log(2)$$
where 
$$T_1 = \left(   \begin{array}{ccccccc}   +0&+1&+1&+1&+1&+1&+1&\cdots \\ +0&-1&+1&-1&+1&-1&+1 \\ +0&+1&-2&+1&+1&-2&+1 \\ +0&-1&+1&-1&+1&-1&+1 \\ +0&+1&+1&+1&-4&+1&+1 \\ +0&-1&-2&-1&+1&+2&+1 \\ +0&+1&+1&+1&+1&+1&-6 \\ \vdots&&&&&&&\ddots \end{array}   \right)$$
$\log(2)$ can be subtracted from the matrix as a sum from the second column:
So again by analytic continuation (not numerically directly):
$$\log(\pi)=\sum\limits_{k=1}^{\infty}\sum\limits_{n=1}^{\infty} \frac{T_2(n,k)}{n^1 \cdot k^0}$$
where 
$$T_2 = \left(   \begin{array}{ccccccc}   +0&+0&+1&+1&+1&+1&+1&\cdots \\ +0&-0&+1&-1&+1&-1&+1 \\ +0&+0&-2&+1&+1&-2&+1 \\ +0&-0&+1&-1&+1&-1&+1 \\ +0&+0&+1&+1&-4&+1&+1 \\ +0&-0&-2&-1&+1&+2&+1 \\ +0&+0&+1&+1&+1&+1&-6 \\ \vdots&&&&&&&\ddots \end{array}   \right)$$
Hope this helps.
