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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!

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  • $\begingroup$ This would depend on what $A_k$ are. $\endgroup$ – Robert Israel Mar 4 at 15:37
  • $\begingroup$ $A_k$ is just a positive constant $\endgroup$ – meff11 Mar 4 at 15:41
  • $\begingroup$ Why does it have a subscript $k$? $\endgroup$ – Robert Israel Mar 4 at 15:47
  • $\begingroup$ 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 $\endgroup$ – meff11 Mar 4 at 15:52
  • $\begingroup$ Then, as Robert Israel said, convergence depends upon the values of the $A_k$. $\endgroup$ – user247327 Mar 4 at 16:07
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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.

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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 $x$.

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