I have a symmetric $n*n$ adjacency circulant matrix with a band structure given as follows:
\begin{bmatrix} 0 &1 &1&.&.&0 \\ 1 &0 &1&1&.&.\\ 1 &1 &.&.&.&.\\ . &1 &.&.&.&1\\ 0 &. &.&.&.&1\\ 0 &. &.&1&1&0\\ \end{bmatrix} The elements in the diagonal take the value zero. There is a band of size $M(n)$ and the elements in that band take the value $1$. I have $\dfrac{M(n)}{n}=0$ as $n \to \infty$.
I'am interested on finding the limiting eigenvalue distribution of this specific matrix.
My first though was to use the following property:
If $C \in \mathbb R^{n \times n}$ is a circulant matrix. Then its eigenvalues have the form (e.g. here [PDF]) $$\lambda_k = \sum_{j = 0}^{n - 1} c_j \exp\left(\frac{2\pi i}{n} k\right), \tag1$$ where $c = (c_0, c_1, \ldots, c_{n - 1})$ is the first row of $C$.
Applying this result I get the following expression for my matrix: $$ \lambda_{k}= \sum\limits_{\substack{k=0}}^{n-1}{t_{k} \exp\left(\frac{2\pi i}{n} k\right)}= \sum\limits_{\substack{k=0}}^{M(N)}{t_{k} \exp\left(\frac{2\pi i}{n} k\right)}+\sum\limits_{\substack{k=n-2M(N)}}^{n-1}{t_{k} \exp\left(\frac{2\pi i}{n} k\right)}\\ $$ From $k=0$ to $k=M(n)$, $t_{k}=1$ otherwise zero. thus, we just keep the first sum of the previous expression.
$$\lambda_{k}=\sum\limits_{\substack{k=0}}^{M(n)}{\exp\left(\frac{2\pi i}{n} k\right)}=\left[ \dfrac{\sin\Big(\frac{\pi k}{n}(2M(n)+1)\big)}{\sin\big(\dfrac{\pi k}{n}\Big)} \right] -1 $$
But when I let $n \to \infty$, I get an indeterminate form and the limit is not computable. Is there any possible trick to handle that kind of matrices.