In this publication, in proof of theorem 5.11 there is transition which I interpret as inequality $$ \sum_{k=0}^\infty 2^{Nk} \exp\left(-\frac {4^k \rho^2} {Dt}\right) \leq C_{N, D} \exp\left(-\frac {\rho^2} {Dt}\right).$$ Here $D>2$ may be chosen, $N>0$ is fixed and $C$ can depend on $D$ and $N$. Additionally we may assume $\rho^2/t \geq 1$, and this expression is treated as free variable. Last inequality is highlighted in the publication, but I don't see why that would be useful; the author talks something about series on the left being majorized by geometric series.
I tried to write some integral inequalities, but I was able to prove the above only for $N \leq 2$ which is far from satisfactory.
How do I prove this inequality?
To start you can use $$ 2^{Nk}\exp\left ( -\frac{4^k\rho^2}{Dt}\right ) = \exp(\ln(2))^{Nk}\exp\left ( -\frac{4^k\rho^2}{Dt}\right ) = \exp \left (\ln(2)Nk - \frac{4^k \rho^2}{Dt} \right ) $$ Now let us pick a $K$ so that for $k \geq K$ we have the bound $$ \ln(2)Nk - \frac{4^k\rho^2}{Dt} \leq -\frac{4^k\rho^2}{2Dt} $$ Such a $K$ can be taken to be $$ K = \max\left \{ 1, \left \lceil \frac{\ln(2\ln(2)ND)}{1-1/e} \right \rceil \right \} $$ The max just deals with the edge case of $N$ being very small. I'll give a detailed proof of this below. Note that this bound $K$ only depends on $N$ and $D$ and it does so in an essentially logarithmic way.
Now let's break the infinite sum into two pieces. \begin{align} \sum_{k=0}^\infty 2^{Nk}\exp\left (-\frac{4^k\rho^2}{Dt} \right ) &= \sum_{k=0}^{K-1} \exp\left ( \ln(2)Nk - \frac{4^k\rho^2}{Dt} \right ) + \sum_{k=K}^\infty \exp\left ( \ln(2)NK - \frac{4^k\rho^2}{Dt} \right ) \\ &\leq \sum_{k=0}^{K-1} \exp\left ( \ln(2)NK - \frac{4^k\rho^2}{Dt} \right ) + \sum_{k=K}^{\infty} \exp\left ( -\frac{4^k\rho^2}{2Dt} \right ). \end{align} The remaining strategy is a three step procedure which is summarized as follows.
Show that the second summation is bounded by a constant multiple of its first term $$\sum_{k=K}^{\infty} \exp\left ( -\frac{4^k\rho^2}{2Dt} \right ) \leq C_1\exp\left ( -\frac{4^K\rho^2}{2Dt} \right )$$ where the constant $C_1$ must be controlled appropriately.
Show that the this expression that controls the second summation is itself controlled by last term of the first summation, that is $$\exp\left ( -\frac{4^K\rho^2}{2Dt} \right ) \leq \exp\left ( \ln(2)N(K-1) - \frac{4^{K-1}\rho^2}{Dt} \right )$$
Show that the first summation is controlled by the required bound $$\sum_{k=0}^{K-1} \exp\left ( \ln(2)NK - \frac{4^k\rho^2}{Dt} \right ) \leq C_{N,D}' \exp\left ( \frac{\rho^2}{Dt} \right )$$ and choose an appropriate $C_{N,D}$ by using $C_{N,D}'$ and the arguments of the first two steps.
I'll do Step 2 first because it is quite simple compared to the others.
Step 2 This step can be shown using some fundamental exponent rules: \begin{align} \exp\left ( -\frac{4^K\rho^2}{2Dt} \right ) &= \exp\left ( -\frac{4^{K-1/2}\rho^2}{Dt} \right ) \\ &\leq \exp\left ( -\frac{4^{K-1}\rho^2}{Dt} \right ) \\ &\leq \exp\left (\ln(2)N(K-1) -\frac{4^{K-1}\rho^2}{Dt} \right ) \end{align}
Step 1 Now we will analyze the second summation. The rough idea is that $\exp\left ( -\frac{4^k\rho^2}{2Dt} \right )$ decays super-exponentially so it is reasonable to expect that in the summation $$\sum_{k=K}^{\infty} \exp\left ( -\frac{4^k\rho^2}{2Dt} \right )$$ only the largest term contributes much. In fact I'm going to use what is probably a pretty loose bound but it is good enough.
The key is to use the inequality $4^{K+x} \geq 4^K + 4x$ which is valid for all $K \geq 1$ and $x \geq 0$. This can be seen by noting that at $x = 0$ they are the same, and the derivative of the left is $4^{K+x}\ln(4)$ while on the right it is just $4$. Using this (and a lot of small manipulations) gives \begin{align} \sum_{k=K}^{\infty} \exp\left ( -\frac{4^k\rho^2}{2Dt} \right ) &= \sum_{i=0}^{\infty} \exp\left ( -\frac{4^{K+i}\rho^2}{2Dt} \right ) \\ &\leq \sum_{i=0}^\infty \exp \left ( -\frac{4^{K}\rho^2}{2Dt} - \frac{4i\rho^2}{2Dt} \right ) \\ &= \exp \left ( -\frac{4^{K}\rho^2}{2Dt} \right ) \sum_{i=0}^\infty \exp \left (- \frac{2\rho^2}{Dt} \right )^i \\ &= \exp \left ( -\frac{4^{K}\rho^2}{2Dt} \right ) \frac{1}{1 - \exp(-2\rho^2/Dt)} \end{align} Where we have used the formula for the sum of a geometric series. I believe that this is probably the part that they were referring to about being majorized by a geometric series. Now using that $\rho^2/t \geq 1$ and the bound above we have \begin{align} \sum_{k=K}^{\infty} \exp\left ( -\frac{4^k\rho^2}{2Dt} \right ) &\leq \exp \left ( -\frac{4^{K}\rho^2}{2Dt} \right ) \frac{1}{1 - \exp(-2\rho^2/Dt)} \\ &\leq \exp \left ( -\frac{4^{K}\rho^2}{2Dt} \right ) \frac{1}{1 - \exp(-2/D)} \end{align} which is the bound we need with $$C_1 = \frac{1}{1 - \exp(-2/D)}$$
Step 3 Overall we have shown so far that \begin{align} \sum_{k=K}^{\infty} \exp\left ( -\frac{4^k\rho^2}{2Dt} \right ) &\leq C_1 \exp\left ( -\frac{4^K\rho^2}{2Dt} \right ) & (\text{Step 1}) \\ &\leq C_1 \exp\left ( \ln(2)N(K-1) - \frac{4^{K-1}\rho^2}{Dt} \right ) & (\text{Step 2}) \\ &\leq C_1 \sum_{k=0}^{K-1} \exp\left ( \ln(2)Nk - \frac{4^{k}\rho^2}{Dt} \right ) \end{align} Now let's work on just this summation. Here we will use a fairly crude bound \begin{align} \sum_{k=0}^{K-1} \exp\left ( \ln(2)Nk - \frac{4^{k}\rho^2}{Dt} \right ) &\leq \sum_{k=0}^{K-1} \exp\left ( \ln(2)Nk - \frac{\rho^2}{Dt} \right ) \\ &= \exp \left (- \frac{\rho^2}{Dt} \right ) \sum_{k=0}^{K-1} 2^{Nk} \\ &\leq \exp \left (- \frac{\rho^2}{Dt} \right ) K2^{N(K-1)} \\ &= C_{N,D}' \exp \left (- \frac{\rho^2}{Dt} \right ) \end{align} where $C_{N,D}' = K2^{N(K-1)}$ which only depends on $N$ and $D$ because $K$ is a function of $N$ and $D$. (Note I got a bit lazy here, one can explicitly compute the finite geometric sum).
Now putting all the pieces together we have \begin{align} &\sum_{k=0}^{\infty} \exp\left ( \ln(2)Nk - \frac{4^{k}\rho^2}{Dt} \right ) \\ &\leq (1+C_1) \sum_{k=0}^{K-1} \exp\left ( \ln(2)Nk - \frac{4^{k}\rho^2}{Dt} \right ) \\ &\leq (1+C_1)C_{N,D}' \exp \left (- \frac{\rho^2}{Dt} \right ) = C_{N,D}\exp \left (- \frac{\rho^2}{Dt} \right ) \end{align} where $$C_{N,D} = (1+C_1)C_{N,D}'$$ which is valid since $C_1$ only depends on $D$.
Retrospective
The general strategy of the proof can be summarized
1.Breaking the summation into two parts. A small $k$ part where the super-exponential decay has yet to dominate and a large $k$ part where the super-exponential decay is all that matters. The trick is to find where to break these two parts.
Doing some analysis to show that in the super-exponential decay, only the first term (the $K$ term) really matters, and that this term is controlled by the $K-1$ term.
More analysis based on the fact that $K$ isn't too large and therefore the small $k$ summation doesn't have too many terms.
The bound on $K$
This is a bit of work
\begin{align} \ln(2)Nk - \frac{4^k\rho^2}{Dt} \leq -\frac{4^k\rho^2}{2Dt} &\iff \ln(2)Nk \leq \frac{4^k\rho^2}{2Dt} \\ &\iff \frac{\ln(2)N2Dt}{\rho^2}k \leq 4^k \end{align} Now let $a = \frac{\ln(2)N2Dt}{\rho^2}$ and $b = 4$. We seek a bound of the form $$ak \leq b^k \iff \ln(a) + \ln(k) \leq \ln(b)k \iff \ln(a) \leq \ln(b)k-\ln(k)$$ Now $\ln(4) > \ln(e) = 1$ and $\ln(k) \leq \frac{k}{e}$ so that $$\ln(k)x - \ln(k) > k - \frac{k}{e} = (1-1/e)k.$$ In particular $$\frac{\ln(a)}{(1-1/e)} < k \implies \ln(a) < (1-1/e)k < \ln(bk) - \ln(k) $$
This shows that for $k > \frac{\ln(a)}{(1-1/e)}$ we have the desired inequality. Plugging back in the definition of $a$ and using the ceiling to get an integer we have a suitable $K$ at $$K = \left \lceil \frac{\ln(2\ln(2)NDt/\rho^2)}{1-1/e} \right \rceil$$ We can remove the $t/\rho^2$ by using the fact that it is bounded by $1$ by assumption which gives $$K = \left \lceil \frac{\ln(2\ln(2)ND)}{1-1/e} \right \rceil$$ and increasing $K$ to the max of this and 1 does no harm either.