Show $\sum_{n=1}^{k} 2^{a_n} = 2^{a_1+1}$ holds for all non negative, monotonic decreasing sequences. Let $(a_n)_{n\in\mathbb{N}}$ be a non negative, monotonic decreasing sequence with each $a$ being a whole number. Prove that there exists a $k$ such that
$$
\sum\limits_{n=1}^{k} 2^{a_n} = 2^{a_1+1}
$$
holds.

This result seems rather obvious and intuitive but I want a rather simple and concise proof, which I was so far unable to find.
Suppose otherwise. Since $2^{a_n}\geq1$, we could take $k$ such that $$\sum_{n=1}^k2^{a_n}&lt;2^{a_1+1}&lt;\sum_{n=1}^{k+1}2^{a_n}\Leftrightarrow$$ $$\sum_{n=1}^k2^{a_n-a_k}&lt;2^{a_1-a_k+1}&lt;\sum_{n=1}^{k+1}2^{a_n-a_k}.$$
Since the numbers in the last inequality are all integers, this must imply that $$\sum_{n=1}^{k+1}2^{a_n-a_k}-\sum_{n=1}^k2^{a_n-a_k}\geq2\Leftrightarrow$$ $$2^{a_{k+1}-a_k}\geq2.$$ However, since $a_{k+1}\leq a_k$, this is absurd. We've reached our contradiction.
If $a_n$ is allowed to be negative this isn't so. Assuming "whole number" means $a_n\ge0$:

Since $2^{a_n}\ge1$ there exists $k$ with  $
\sum\limits_{n=1}^{k} 2^{a_n} \ge 2^{a_1+1}.
$ Choose $k$ as small as possible satisfying this inequality, so $$
\sum\limits_{n=1}^{k} 2^{a_n} \ge 2^{a_1+1}
$$while $$
\sum\limits_{n=1}^{k-1} 2^{a_n} &lt; 2^{a_1+1}.
$$

Now say $$2^{a_1+1}=Ka_{k}.$$Since $
\sum\limits_{n=1}^{k-1} 2^{a_n} 
$is a multiple of $a_k$ we have $$
\sum\limits_{n=1}^{k-1} 2^{a_n} \le (K-1)a_k.
$$It follows that in fact $\sum\limits_{n=1}^{k-1} 2^{a_n} = (K-1)a_k$, since otherwise $\sum\limits_{n=1}^{k-1} 2^{a_n} \le (K-2)a_k$, which would imply $$\sum\limits_{n=1}^{k-1} 2^{a_n} \le (K-1)a_k&lt;2^{a_1+1},$$contradiction. So $$\sum\limits_{n=1}^{k} 2^{a_n} = Ka_k=2^{a_1+1}.$$