Gerschgorin's circle theorem gives the upper bound $2(a+1)$ for the eigenvalues. This bound should be sharp if I'm not mistaken. I think the family of $n\times n$ circulant matrices $$\begin{pmatrix} 2 & -1 & 0 & \cdots & 0 & -1 \\ -1 & 2 & -1 & 0 & \cdots & 0 \\ \vdots & & \ddots & & & \vdots \\ 0 & \cdots & 0 & -1 & 2 & -1 \\ -1 & 0 & \cdots & 0 & -1 & 2 \end{pmatrix}$$ should have maximal eigenvalue arbitrarly close to 2 for $n\gg 0$. If I didn't make any mistakes the maximal eigenvalue of this matrix is $2-2\cos(\frac{2\pi}{n})$.

EDIT: And just as I finished writing the first part of this post I remembered the Perron-Frobenius theorem. Considere the matrices with constant diagonal $a+1$ and exactly $a+1$ times an entry equal to +1 off the diagonal in each row. These are positive matrices and have constant row sum $2(a+1)$. Therefore this row sum is an eigenvalue with eigenvector $(1,1,\cdots,1)^T$. It coincides with the $\infty$-operator norm (=row sum norm) of the matrix and is therefore equal to the spectral radius. These matrices therefore attain the upper bound exactly.

Gerschgorin's circle theorem gives the upper bound $2(a+1)$ for the eigenvalues. This bound should be sharp if I'm not mistaken. I think the family of $n\times n$ circulant matrices $$\begin{pmatrix} 2 & -1 & 0 & \cdots & 0 & -1 \\ -1 & 2 & -1 & 0 & \cdots & 0 \\ \vdots & & \ddots & & & \vdots \\ 0 & \cdots & 0 & -1 & 2 & -1 \\ -1 & 0 & \cdots & 0 & -1 & 2 \end{pmatrix}$$ should have maximal eigenvalue arbitrarly close to $2\cdot (1+1)$ (here $a=1$) for $n\gg 0$. If I didn't make any mistakes the eigenvalues of this matrix are $2-2\cos(\frac{2\pi j}{n})$ for $j=0,1,\ldots,n-1$. For $j=\lfloor \frac{n}{2} \rfloor$ and $j=\lceil \frac{n}{2} \rceil$ this is maximal and arbitrary close to 4, even equal to 4 if $2\mid n$.

EDIT: And just as I finished writing the first part of this post I remembered the Perron-Frobenius theorem. Considere the matrices with constant diagonal $a+1$ and exactly $a+1$ times an entry equal to +1 off the diagonal in each row. These are positive matrices and have constant row sum $2(a+1)$. Therefore this row sum is an eigenvalue with eigenvector $(1,1,\cdots,1)^T$. It coincides with the $\infty$-operator norm (=row sum norm) of the matrix and is therefore equal to the spectral radius. These matrices therefore attain the upper bound exactly.

Gerschgorin's circle theorem gives the upper bound $2(a+1)$ for the eigenvalues. This bound should be sharp if I'm not mistaken. I think the family of $n\times n$ circulant matrices $$\begin{pmatrix} 2 & -1 & 0 & \cdots & 0 & -1 \\ -1 & 2 & -1 & 0 & \cdots & 0 \\ \vdots & & \ddots & & & \vdots \\ 0 & \cdots & 0 & -1 & 2 & -1 \\ -1 & 0 & \cdots & 0 & -1 & 2 \end{pmatrix}$$ should have maximal eigenvalue arbitrarly close to 2 for $n\gg 0$. If I didn't make any mistakes the maximal eigenvalue of this matrix is $2-2\cos(\frac{2\pi}{n})$.

Gerschgorin's circle theorem gives the upper bound $2(a+1)$ for the eigenvalues. This bound should be sharp if I'm not mistaken. I think the family of $n\times n$ circulant matrices $$\begin{pmatrix} 2 & -1 & 0 & \cdots & 0 & -1 \\ -1 & 2 & -1 & 0 & \cdots & 0 \\ \vdots & & \ddots & & & \vdots \\ 0 & \cdots & 0 & -1 & 2 & -1 \\ -1 & 0 & \cdots & 0 & -1 & 2 \end{pmatrix}$$ should have maximal eigenvalue arbitrarly close to 2 for $n\gg 0$. If I didn't make any mistakes the maximal eigenvalue of this matrix is $2-2\cos(\frac{2\pi}{n})$.

EDIT: And just as I finished writing the first part of this post I remembered the Perron-Frobenius theorem. Considere the matrices with constant diagonal $a+1$ and exactly $a+1$ times an entry equal to +1 off the diagonal in each row. These are positive matrices and have constant row sum $2(a+1)$. Therefore this row sum is an eigenvalue with eigenvector $(1,1,\cdots,1)^T$. It coincides with the $\infty$-operator norm (=row sum norm) of the matrix and is therefore equal to the spectral radius. These matrices therefore attain the upper bound exactly.