= Solution
Write $h=\Delta x$, $k=\Delta t$ and $\delta_{xx}u_i=(u_{i+1}-2u_i+u_{i-1})/h^2$. Insert a smooth exact <heat equation> solution into the <Crank-Nicolson method>. Expanding about the midpoint $(x_i,t_n+k/2)$ gives
$$
\frac{u(t+k/2)-u(t-k/2)}k=u_t+\frac{k^2}{24}u_{ttt}+O(k^4),
$$
$$
\frac D2[\delta_{xx}u(t-k/2)+\delta_{xx}u(t+k/2)]
=D\left[u_{xx}+\frac{h^2}{12}u_{xxxx}+\frac{k^2}{8}u_{xxtt}\right]+O(h^4+k^4+h^2k^2).
$$
Thus the residual, defined as left side minus right side, is
$$
\mathcal T=\frac{k^2}{24}u_{ttt}-\frac{Dk^2}{8}u_{xxtt}-\frac{Dh^2}{12}u_{xxxx}+O(h^4+k^4+h^2k^2).
$$
Since $u_t=Du_{xx}$ implies $u_{ttt}=Du_{xxtt}$,
$$
\boxed{\mathcal T=-\frac{k^2}{12}u_{ttt}-\frac{Dh^2}{12}u_{xxxx}+O(h^4+k^4+h^2k^2).}
$$
This proves second-order <numerical consistency> in both space and time. About the old time level, uncombined Taylor remainders can include $O(k^3)$; the midpoint expression displays the symmetric leading errors most cleanly. Multiplying by $k$ gives the one-step defect $O(k^3+kh^2)$. Stable evolution with suitable initial and boundary data then gives global error $O(k^2+h^2)$; consistency alone does not prove global convergence.
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