Limit of Square Root of a Tridiagonal Matrix Consider a tridiagonal $n\times n$ matrix with the following format 
$$ \mathbf A = \begin{bmatrix}
a & b &        & \phantom{\ddots} & \phantom{\ddots}    \\
b & a & b      & \phantom{\ddots} &     \\
  & b & a      & \ddots           &     \\
  &   & \ddots & \ddots           & b   \\
\phantom{\ddots}  & \phantom{\ddots}  &  \phantom{\ddots}      & b                & a   \\
\end{bmatrix}.$$
This matrix has eigenvalues equal to $$\lambda_k = a-2b\cos\left(\frac{k\pi}{1+n}\right),\;\text{ for }\;k\in\{1,\dots,n\}.$$  It follows that $$\lim_{n\to\infty}\lambda_k=a-2b.$$
It is also symmetric which mean it can be diagonalized into $$\mathbf A= \mathbf Q \mathbf \Lambda \mathbf Q^T,\;\text{ so that  }\;\mathbf A^{1/2}= \mathbf Q \mathbf \Lambda^{1/2} \mathbf Q^T.$$ 
Is it possible to obtain an analitical formula for the elements of $\lim_{n\to\infty}\mathbf A^{1/2}$?
Edit: This paper on tridiagonal Toeplitz matrices appears to be helpful.
 A: The eigen-decomposition of an $n\times n$ real symmetric tridiagonal Toeplitz matrix is well-known: $\mathbf A$ can be orthogonally diagonalised as
$$
\mathbf A = Q\operatorname{diag}(\lambda_1,\ldots,\lambda_n)\,Q^\top,
$$
where $\lambda_i = a+2b\cos\left(\frac{i\pi}{n+1}\right)$ and $q_{ij}=\sqrt{\frac2{n+1}}\sin\left(\frac{ij\pi}{n+1}\right)$. So, the entries of $\mathbf B=Q\operatorname{diag}(\sqrt{\lambda_1},\ldots,\sqrt{\lambda_n})\,Q^\top$ are given by
$$
b_{ij}=\sum_{k=1}^n\sqrt{\lambda_k}\,q_{ik}\,q_{jk}
=\frac{2}{n+1}\sum_{k=1}^n\sqrt{a+2b\cos\left(\frac{k\pi}{n+1}\right)}\sin\left(\frac{ik\pi}{n+1}\right)\sin\left(\frac{jk\pi}{n+1}\right).
$$
Therefore, for any two given indices $i$ and $j$,
$$
\lim_{n\to\infty}b_{ij}
=2\int_0^1\sqrt{a+2b\cos(\pi x)}\,\sin(i\pi x)\,\sin(j\pi x)\,dx.
$$
A: I just read the very good @user1551's post. I would like to make some remarks.

*

*Necessarily, each matrix $(A_n)$ of the sequence must be $\geq 0$; that implies that the eigenvalues $(\lambda_k)$ must be $\geq 0$; finally, the condition is $a\geq 2|b|$.


*The OP's remark $\lim \lambda_k=a-2b$ is non-sense, because (for example) the $\lambda_5$ of $A_{10}$ has no relation with the $\lambda_5$ of $A_{20}$.


*The sequence $(A_n)$ is Toeplitz; let $t_i=a_{i+1,1},t_0=a_{i,i},t_{-i}=a_{1,i+1}$. Here $t_0=a,t_1=t_{-1}=b$ and the other are $0$ (the non-zero part is finite).
In general, we associate the Fourier series $f(\lambda)=\sum_kt_ke^{ik\lambda}$; then $t_k=1/2\pi\int_0^{2\pi}f(\lambda) e^{-ik\lambda}d\lambda$ and it is assumed that $\sum_k |t_k|<\infty$ or $\sum_k t_k^2<\infty$. We denote by $T_n(f)$ the associated Toeplitz.
Here $f(\lambda)=a+2b\cos(\lambda)$.


*user1551 found $\lim \sqrt{A_n}=B$ in the sense of pointwise convergence. Note also that his proof shows that for "any" continuous function $g$,
$[\lim g(A_n)]_{i,j}=2\int_0^1g(a+2b\cos(\pi x))\,\sin(i\pi x)\,\sin(j\pi x)\,dx$.
Yet, such a limit is no more Toeplitz. Yet, we can find a Toeplitz limit in the following sense: two sequences $(A_n),(B_n)$ are said to be asymptotically equivalent (AE) when  they are bounded and when $tr((A_n-B_n)^T(A_n-B_n))=o(n)$. Note that the terms $A_n[i,j],B_n[i,j]$ may be very different when $i+j$ is small but they are close when $i+j$ is large.
In fact, one has: let $h,g$ be $2$ "good" Fourier series. Then $T_n(hg)$ and $T_n(h)T_n(g)$ or $T_n(g)T_n(h)$ are AE; moreover $T_n(g^{-1})$ and $T_n(g)^{-1}$ are AE.
Here we consider $T_n(\sqrt{f})=T_n(\sqrt{a+2b\cos(\lambda)})$.
Its associated sequence $t'_k$ is given by $t'_k=1/\pi\int_0^{\pi}\sqrt{a+2b\cos\lambda}\cos(k\lambda)d\lambda$. It is easy to see that the series $|t'_k|$ or ${t'_k}^2$ converges; with a little bit of work, we should be able to show that the sequences $T_n(\sqrt{f})$ and $(\sqrt{A_n})$ are AE.
