Let $0<r<1$. Why $\sum r^{n!}$, $\sum r^{(2^n)}$ diverges as $r$ tends to $1$? It seems obvious since the limit of summand in $r$ is not zero, but the limit of summand in $n$ is zero.
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Let $N$ be arbitrary but fixed; to show divergence of $\sum_n r^{n!}$ as $r \to 1$ we need to be able to find an $r$ such that $\sum_n r^{n!} > N$.
As the comment by David Mitra hints at, we can already arrange for the first $N+1$ terms in the summation to exceed $N$. For, take:$$r = \sqrt[(N+1)!]{\frac N{N+1}}$$
Then it is not too hard to see that if $n \le N+1$, the $n$th summand exceeds $\dfrac N{N+1}$ and therefore that the entire sum must exceed $N$.
The same argument works, mutatis mutandis, for other powers than $n!$.
Lord_Farin
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