I need to prove that $u:\mathbb{R}\times(-\frac{\pi}{2},\frac{\pi}{2})\rightarrow\mathbb{R}$
$u(x,y)=\sum_{n \ \text{ is odd}}\cos(ny)e^{n(x-n)}$
is harmonic. I have no idea which theorem or result to use.
I need to prove that $u:\mathbb{R}\times(-\frac{\pi}{2},\frac{\pi}{2})\rightarrow\mathbb{R}$
$u(x,y)=\sum_{n \ \text{ is odd}}\cos(ny)e^{n(x-n)}$
is harmonic. I have no idea which theorem or result to use.
Im gonna show that there exists $n_{0}\in\mathbb{N}$ such that for $|x|\leq\delta$ and $\forall \ y$, we have $\Big|\sum_{k=n_{0}}^{\infty}\cos((2k-1)y)e^{(2k-1)(x-(2k-1))}\Big|<\epsilon$ for some $\epsilon$.
In fact we have
\begin{eqnarray} \Big|\sum_{k=n_{0}}^{\infty}\cos((2k-1)y)e^{(2k-1)(x-(2k-1))}\Big| &\leq& \sum_{k=n_{0}}^{\infty}\Big|\cos((2k-1)y)e^{(2k-1)(x-(2k-1))}\Big| \nonumber \\ &\leq& \sum_{k=n_{0}}^{\infty}\Big|e^{(2k-1)(x-(2k-1))}\Big| \nonumber \end{eqnarray}
Now take $x=\delta$. It is easy to see that there exists $n_{0}$ such that the sum above is less than $\epsilon$ (because exponential decay too fast). If $x<\delta$, as exponential is increasing, you still have the same thing.
With a little adaptation of the argument above, you can show that the convergence is uniform and then you can use the result posted by Nate Eldredge.
The function has to satisfy the equation $ u_{xx} + u_{yy}=0$.