I have a last minute homework problem that I'm struggling with, and ANY help is greatly needed (and extremely appreciated)!
Problem (exactly how it is given): Recall that the series solution to the Dirichlet heat equation on $0\leq x\leq\ell$ is $u(t,x)={\displaystyle{\sum\limits_{n=1}^{+\infty}}}A_{n}e^{-\big(\frac{n\pi}{\ell}\big)^{2}kt}\sin\bigg(\dfrac{n\pi x}{\ell}\bigg)$ and $\phi(x)={\displaystyle{\sum\limits_{n=1}^{+\infty}}A_{n}\sin\bigg(\dfrac{n\pi x}{\ell}\bigg)}$. Assume that the coefficients $A_{n}$ are absolutely summable, i.e. ${\displaystyle{\sum\limits_{n=1}^{+\infty}}|A_{n}|}<+\infty$. Prove that $u(t,x)\rightarrow 0$ as $t\rightarrow +\infty$.
Additional Remarks: I know the exponential term converges to 0 as $t\rightarrow +\infty$ in the series solution, and for any $\alpha\in\mathbb{R}$ the $|\sin(\alpha)|\leq 1$. However, I keep getting tangled when trying to use the precise definition of the limit of $u$ as $t\rightarrow+\infty$ while keeping $x$ fixed couple with the absolute summability of the $A_{n}$. I figured there must be an easy way to do this problem since we are mainly trying to show $\lim\limits_{t\rightarrow+\infty}\!\!u(x,t)=u(x,+\infty)=0$, and maybe I'm making things to hard in my preliminary work, but any help is GREATLY appreciated!
Observe that $e^{-(n\pi/\ell)^2kt}$ decreases with $n$. Then $$ |u(t,x)|\le\sum_{n=1}^{\infty}|A_{n}|\,e^{-(\tfrac{n\pi}{\ell})^{2}kt}\,\Bigl|\sin\Bigl(\dfrac{n\pi x}{\ell}\Bigr)\Bigr|\le e^{-(\tfrac{\pi}{\ell})^{2}kt}\sum_{n=1}^{\infty}|A_{n}|. $$