We have the following problem
$$ \begin{cases} u \in \mathcal{C} ^1 ([0, \pi] \times [0, \infty[) \cap \mathcal{C}^2 ([0, \pi] \; \times \; ]0, \infty[) \\ u_{xx} + u_{yy} = 0 \quad \forall (x,y) \in\; ]0, \pi[ \; \times \; ]0, \infty[ \\ u(0, y) = u(\pi, y) = 0 \quad \forall y \in\;]0, \infty[ \\ u(x,0)=f(x), \quad u_y(x,0)=0 \quad \forall x \in \; ]0, \pi[ \end{cases}$$
where $f\in\mathcal{C} ^1 ([0,\pi])$.
I have already shown that the formal solution is
$$ u(x,y)=\sum_{n=1}^\infty \beta_n[f] \cosh (ny) \sin (nx), \qquad \beta_n[f] = \frac{2}{\pi} \int_0 ^\pi f(s) \sin(ns) \, ds $$
Now I want to find $f\in\mathcal{C} ^1 ([0,\pi])$ and $ (x,y) \in [0, \pi] \times [0, \infty[ $ such that
$$ \sum_{n=1}^\infty \beta_n[f] \cosh (ny) \sin (nx) $$
doesn't converge.
Since $\sum_{n=1} ^\infty \frac{\sin (nx)}{n^2}$ is convergent for all $x\in \mathbb{R}$ we can take
$$ f(x) = \sum_{n=1} ^\infty \frac{\sin (nx)}{n^2} $$
So we have $ \beta_n[f] = \frac{1}{n^2}$ and
$$ u(x,y) = \sum_{n=1} ^\infty \frac{\cosh (ny) \sin (nx)}{n^2} $$
But, for $(x,y) \in \; ]0, \pi [ \; \times \; ]0, \infty[$
$$ \nexists \lim_{n\to \infty} \frac{\cosh (ny) \sin (nx)}{n^2} $$
So the series doesn't converge.