Accumulation of nodes is not possible for non trivial wavefunction. If there exists a limit point of nodes, then, the value of ${\displaystyle \psi '(p)=0}$ for the limit point ${\displaystyle p}$ but this corresponds to trivial solution having initial values ${\displaystyle \psi (p)=\psi '(p)=0}$.

From Schrodinger's equations:

${\displaystyle (E_{i}-V(x,t)){\Psi _{i}(x,t)}=-{\frac {\hbar ^{2}}{2m}}{\frac {\partial ^{2}\Psi _{i}(x,t)}{\partial x^{2}}}}$

Multiplying with ${\displaystyle \Psi _{j}(x,t)}$ and subtracting equations of ${\displaystyle i=1\,,j=2}$ and ${\displaystyle i=2\,,j=1}$

${\displaystyle (\psi _{1}''\psi _{2}-\psi _{2}''\psi _{1})=(\psi _{1}^{\prime }\psi _{2}-\psi _{2}^{\prime }\psi _{1})^{\prime }=(E_{2}-E_{1})\psi _{1}\psi _{2}}$

If ${\displaystyle x_{1}}$ and ${\displaystyle x_{2}}$ are consecutive nodes of ${\displaystyle \psi _{1}}$, assuming it to be positive in the region, implies ${\displaystyle \psi _{1}^{\prime }(x_{1})>0}$ and ${\displaystyle \psi _{1}^{\prime }(x_{2})<0}$. Computing the following integration:

${\displaystyle \psi _{1}^{\prime }(x_{2})\psi _{2}(x_{2})-\psi _{1}^{\prime }(x_{1})\psi _{2}(x_{1})=(E_{2}-E_{1}){\int _{x_{1}}^{x_{2}}\!dx\,\psi _{1}(x)\psi _{2}(x)}}$

If ${\displaystyle E_{2}>E_{1}}$, ${\displaystyle \psi _{2}}$ cannot be entirely positive or entirely negative in the region and must therefore have node within the consecutive nodes of ${\displaystyle \psi _{1}}$. Note that this is independent on assuming sign of ${\displaystyle \psi _{1}}$ in this region. Thus wavefunctions with different energy levels always have different number of nodes or conversely wavefunctions with same number of nodes cannot have different energies. Thus in a 1D bound system where degeneracy is not allowed, all energy eigenfunctions differ by the number of nodes and when arranged in increasing order of energies, give increasing number of node.