Does $\sum\limits_{k=1}^{\infty}A_ke^{-k^2t}\sin(kx)$ converge for t >0?

Show that the series solution converges for each $$t > 0$$

$$\sum\limits_{k=1}^{\infty}A_ke^{-k^2t}\sin(kx)$$

I'm really rusty on the different convergence tests. Any help would be appreciated!

• This would depend on what $A_k$ are. – Robert Israel Mar 4 at 15:37
• $A_k$ is just a positive constant – meff11 Mar 4 at 15:41
• Why does it have a subscript $k$? – Robert Israel Mar 4 at 15:47
• This series is a solution to a PDE. Each $A_ke^{-k^2t}sin(kx)$ is a solution to the PDE for $k=1, 2,...$. Thus by superposition, $\sum A_ke^{-k^2t}sin(kx)$ is also a solution – meff11 Mar 4 at 15:52
• Then, as Robert Israel said, convergence depends upon the values of the $A_k$. – user247327 Mar 4 at 16:07

If you intend to define the $$A_k$$s as Fourier series coeffs of a well-defined initial condition, then $$A_k$$s are bounded. Under that circumstance, we can write $$\left|\sum\limits_{k=1}^{\infty}A_ke^{-k^2t}\sin(kx)\right|{\le \sum\limits_{k=1}^{\infty}|A_k|\cdot e^{-k^2t}\cdot|\sin(kx)|\\\le \sum\limits_{k=1}^{\infty}|A_k|\cdot e^{-k^2t}\\\le \sum\limits_{k=1}^{\infty}\text{Constant}\cdot e^{-k^2t}\\=\text{Constant}\cdot \sum\limits_{k=1}^{\infty}e^{-k^2t}\\=\text{Bounded}}$$therefore the series converges.
Typically you'll get the $$A_k$$ as Fourier series coefficients for the initial condition at $$t=0$$. In particular, $$A_k$$ will be bounded. Then you can use a comparison test with a geometric series to show that your series converges for all $$t > 0$$ and real $$x$$.