Random walk on Non-negative integers (Invariant Dist.) I wanted to confirm my answer to the following: 

Given a random walk on the non-negative integers, beginning at 0 and provided transition probabilities of $\frac{1}{n+1}$ to the right and $\frac{n}{n+1}$ to the left at some arbitrary $\ n$ we have a probability transition matrix as follows. 

$$PTM=\begin{bmatrix}
0 & 1 & 0 & 0 & 0 & \cdots \\ 
1/2 & 0 & 1/2 & 0 & 0 & \cdots\\ 
0 & 2/3 & 0 & 1/3 & 0 & \cdots\\ 
0 & 0 & 3/4 & 0 & 1/4 & \cdots\\ 
0 & 0 & 0 & 4/5 & 0 & \cdots \\ 
\vdots  & \vdots & \vdots & \vdots & \vdots & \ddots 
\end{bmatrix}$$
To determine the invariant distribution, we must find some $\mu$ such that $\mu P=\mu$. Provided that we assume our initial step for the walk is distributed according to $W_0 \sim \mu$, we have that the invariant distribution can be calculated as follows:
$$\frac{1}{2} \mu_1 + \sum_{n=1}^{\infty} \mu_{n-1}\frac{1}{n} + \mu_{n+1}\frac{n+1}{n+2}=1$$
When calculating the above expression per term I create the following expressions:
$$\frac{1}{2} \mu_1 = \mu_0$$
$$\mu_0 + \frac{2}{3} \mu_2 = \mu_1$$
$$\vdots$$
$$\mu_{n-1}\frac{1}{n} + \mu_{n+1}\frac{n+1}{n+2} = \mu_n$$
Solving the above expression I have the following invariant distribution:
$$\mu = \left \langle \mu_0 \ 2\mu_0 \ \frac{3}{2}\mu_0 \ \frac{2}{3}\mu_0 \ \frac{5}{24}\mu_0 \ \frac{1}{20}\mu_0 \ 0 \cdots \right \rangle$$

This is the right approach and the answer (up to the 0 in the last component in your last equation, which seems to be a typo) is also correct. The general form of the distribution is $\mu_n = \frac{n+1}{n!} \mu_0$, which you can prove by induction. 
To make this a probability distribution, choose $\mu_0 = \frac{1}{2e}$. Then all terms sum to 1.
