In our lecture we discussed the solutions to \begin{cases} u_t - u_{xx} = 0, & x \in (0,1) \\ u_x(0,t) = u_x(1,t) = 0, & t > 0 \\ u_x(x,0) = u_0(x) \in L^2(0,1) \end{cases} and found $$ u(x,t) = \frac{B_0}{2} + \sum_{k = 1}^{\infty} e^{- \pi^2 k^2 t} B_k \cos(\pi k x), $$ where the $B_k = \int_{0}^{1} u_0(x) \cos(\pi k x) dx$.
To find the asymptotic behaviour we examined $$ K := \left| u(t,x) - \frac{B_0}{2} \right| = \left| \sum_{k = 1}^{\infty} e^{- \pi^2 k^2 t} B_k \underbrace{\cos(\pi k x)}_{\le 1} \right| \overset{(\star)}{\le} \left( \sum_{k = 1}^{\infty} B^2_k \right)^{\frac{1}{2}} \left( \sum_{k = 1}^{\infty} e^{- \pi^2 k^2 t} \right)^{\frac{1}{2}}, $$ using Cauchy-Schwartz in $(\star)$.
By Parsevals identity we concluded $\sum_{k = 1}^{\infty} B^2_k = \| u_0 \|_{L^2(0,1)}$. Until this point everything is clear.
But now we concluded $$ K \le \| u_0 \|_{L^2} \left( \sum_{k = 1}^{\infty} \left(e^{- t^2} \right)^{k} \right)^{\frac{1}{2}} = \| u_0 \|_{L^2} \left( 1 - \frac{1}{1 - e^{-t^2}} \right) \xrightarrow{t \to \infty} 0. $$ There seem to be many mistakes, beginning with the absence of $\pi^2$ and $\| u_0 \|$ instead of $\| u_0 \|^{\frac{1}{2}}$, which I can't explain, but which are probably typos. My question really concerns the second sum. I know that for $|r| < 1$ we have $\sum_{k = 1}^{\infty} r^k = \frac{r}{r - 1}$ and that $\sum_{k = 1}^{\infty} r^{(k^2)}$ can only be written with the complicated Jacobi Theta functions, which our instructor probably didn't have in mind.
How can this sum be estimated instead?
Update 1 Plotting with WolframAlpha reveals that probably $$ \tag{$\star$} \sum_{k = 1}^{\infty} e^{- \pi^2 k^2 t} \le -\left(1 - \frac{1}{1 - e^{- \pi^2 t}}\right) = \sum_{k = 1}^{\infty} \left(e^{- \pi^2 t}\right)^k, $$ indicating that the above inequality maybe correct up to the sign and the typos mentioned above. How would one show ($\star$)?
This is because for $t>0$ and $k\in \Bbb N$, we can write $$\pi^2k^2t\ge \pi^2 kt$$and since the function $g(w)=e^{-w}$ is strictly decreasing over $\Bbb R$, we obtain$$e^{-\pi^2k^2t}\le e^{-\pi^2kt}$$which yields to what we wanted to show $\blacksquare$