Trust me when I tell you this is not homework. Can you suggest an solid argument to prove $$ S_n = \frac{1}{3^25^27^2\cdots (2n-1)^2}\sum_{m=3}^{\infty} 2^{(2n)m}e^{-2^{m/2}} $$ diverges as $n \rightarrow \infty$ ?
I have calculated $S_1,S_2,S_3,S_4$ and $S_5$ $$ S_1 = \sum_{m=3}^{\infty} 2^{2m}e^{-2^{m/2}} = 13.63 $$ $$ S_2 = \frac{1}{3^2}\sum_{m=3}^{\infty} 2^{4m}e^{-2^{m/2}} = 1611.5$$ $$ S_3 = \frac{1}{3^25^2}\sum_{m=3}^{\infty} 2^{6m}e^{-2^{m/2}} = 511684$$ $$ S_4 = \frac{1}{3^25^27^2}\sum_{m=3}^{\infty} 2^{8m}e^{-2^{m/2}} = 3.42 \times 10^8 $$ $$ S_5 = \frac{1}{3^25^27^29^2}\sum_{m=3}^{\infty} 2^{10m}e^{-2^{m/2}} = \frac{3.513 \times 10^{17}}{3^25^27^29^2} = 3.934 \times 10^{11} $$ and I am convinced it diverges, however, I'd like a more solid argument to rest my conjecture $S_n \rightarrow \infty$ as $n \rightarrow \infty$.
Thanks in advance. I will award 200pts bounty for a quality answer.
This is detailing an observation made by Did in the comments.
Note that, for $n\geq 3$, $$ S_n = \frac{1}{(3\cdot 5 \cdot 7\cdots(2n-1))^2}\sum_{m=3}^\infty 2^{2nm} e^{-2^{m/2}} = \left(\frac{2^n n!}{2n!}\right)^2\sum_{m=3}^\infty 2^{2nm} e^{-2^{m/2}} $$ hence $$S_n\geq \left(\frac{2^n n!}{2n!}\right)^2 \max_{m\geq 3}2^{2nm} e^{-2^{m/2}}\tag{1} $$ This (possibly quite loose) lower bound suggests to choose $m$ as close as possible to the maximizer of the function $f_n\colon[3,\infty)\to\mathbb{R}$ defined by $$ f_n(x) = e^{2n x\ln 2-2^{x/2}}\,. $$ Maximizing $f_n$ is equivalent to maximizing $\ln f_n$, which is done for $x$ such that $2^{x/2} = 4n$. Let $$2\log_2(4n)\leq m_n\stackrel{\rm def}{=} \lceil 2\log_2(4n)\rceil < 2\log_2(4n)+2$$ so that $$ 4n \leq 2^{m_n/2}< 8n\tag{2} $$ Then, $$ S_n \geq \left(\frac{2^n n!}{2n!}\right)^2 f_n(m_n) \geq \left(\frac{2^n n!}{2n!}\right)^2 2^{4n \log_2(4n)}e^{-8n} \stackrel{\rm def}{=} T_n\tag{3} $$ Now, we rely on Stirling's approximation to bound the first factor of $T_n$: $$ T_n \operatorname*{\sim}_{n\to\infty} \frac{1}{2}\cdot\frac{e^{2n}}{2^{2n}n^{2n}} \cdot 2^{4n \log_2(4n)}e^{-8n} = 2^{4n\log_2 n+O(n)} = n^{4n+o(n)} \xrightarrow[n\to\infty]{} \infty \tag{4} $$ By (3) and (4), we get finally $$ \boxed{\lim_{n\to\infty}S_n = \infty} $$