Here,
$$T_n=\begin{pmatrix} 1&&&&&&\\ \frac{n-1}{n}&0&\frac{1}{n}&&&&\\ &\frac{n-2}{n}&0&\frac{2}{n}&&&\\ &&\ddots&\ddots&\ddots&&\\ &&&\frac{2}{n}&0&\frac{n-2}{n}&\\ &&&&\frac{1}{n}&0&\frac{n-1}{n}\\ &&&&&&1 \end{pmatrix}_{(n+1)\times(n+1)}$$
Now what is $\lim_{k\to\infty}T_n^k$?
Note that this chain doesn't have a stationary distribution because of the absorption states which make other states transient (and thus the eigenvalue 1 has multiple corresponding eigenvectors). Experiment shows that the limit has the form of $$T_n^{\infty}=(\mathbf{v,0,\cdots,0,e-v}),$$ where $\mathbf{v}$ is an eigenvector of eigenvalue 1 and can be calculated via Mathematica clause
TransitionMatrix[n_] := Module[{band = Table[k/n, {k, 0, n - 1}]},
SparseArray[{Band[{1, 2}] -> band, Band[{1, 1}] -> {1},
Band[{n + 1, n + 1}] -> {1},
Band[{2, 1}] -> Reverse[band]}]]
NonstationaryDistribution[n_] :=
#/#[[1]]&[Eigenvectors[TransitionMatrix[n]][[2]]]
and $\mathbf{e}$ is a vector of all 1's with a proper length.
Can anyone give some hints? Thank you~