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

Question

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!

Answer 1

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 $x$.

Answer 2

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.