= Solution
The <Householder-John theorem> states: if $A=M-N$ is <Hermitian matrix>[Hermitian] positive definite and $M^*+N$ is also Hermitian positive definite, then $M$ is invertible and
$$
\rho(M^{-1}N)<1.
$$
Now let $A=D+L+L^T$ be real symmetric positive definite and tridiagonal. For the Jacobi splitting, $M=D$ and $N=-(L+L^T)$. Positive definiteness of $A$ implies every diagonal entry is positive, so $D$ is positive definite. Introduce the alternating-sign <diagonal matrix>
$$
S=\operatorname{diag}(1,-1,1,-1,\ldots).
$$
Tridiagonality gives
$$
SAS=D-L-L^T=M^*+N.
$$
Since $S$ is orthogonal, $SAS$ is positive definite whenever $A$ is. Both hypotheses of the Householder-John theorem hold, and hence the Jacobi iteration matrix satisfies $\rho(M^{-1}N)<1$. Therefore the <Jacobi convergence for a symmetric positive-definite tridiagonal matrix> follows.
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