If $p$ and $q$ are coprime positive integers s.t. $frac{p}{q}=sum_{k=0}^{100}frac1{3^{2^k}+1}$, what is the smallest prime factor of $p$?

Question

If the sum $$S=frac14+frac1{10}+frac1{82}+frac1{6562}+cdots+frac1{3^{2^{100}}+1}$$
is expressed in the form $frac pq,$ where $p,qinmathbb N$ and $gcd(p, q) =1.$ Then what is smallest prime factor of $p$ ?

We have: $$S=sum_{k=0}^{100} frac1{3^{2^k}+1}.$$ Please give me some hints to evaluate such kind of sums, in general.
Remark.

Since $4$ divides the denominator of the zeroth term ($frac14$) of $S$, but does not divide the denominator of any other term, we can see that $2nmid p$.

Note that each term of $S$ is $equiv 1pmod{3}$.  Therefore, $Sequiv 101cdot 1equiv 2pmod{3}$, so $3nmid p$.

Since $5$ divides the denominator of the first term ($frac1{10}$) of $S$, but not the denominator of any other term, we conclude that $5nmid p$.

WhatsUp · Answer

I'm not sure what kind of problem this is.
It seems that the problem is not intended to be done by hand.
If we denote by $S(n)$ the number $displaystylesum_{k = 0}^n frac 1{3^{2^k} + 1}$, then the first several values of the numerator of $S(n)$ look like this:
1
7
3^3 * 11
974867
20982415713197
3 * 6480139987906036648979676749
13 * 25220504737903 * 1202418613506277 * 84660948985522106511557529679
149 * 883 * 126001 * 11868766710884224982021663692780373317124689104200960317897970407656906279023556512105818421377935790975902821

Of course, it is easy to show that $3$ divides the numerator if $n equiv 2 pmod 3$. This would be a much more reasonable exercise in elementary number theory.
However we have here $n = 100$. This leads to something without a pattern.
With the help of some computer algebra system, I am able to find that the smallest prime factor of the numerator is $37$. This is done by checking all primes up to $37$ one by one. So it is not really doable with paper and pencil.

If $p$ and $q$ are coprime positive integers s.t. $frac{p}{q}=sum_{k=0}^{100}frac1{3^{2^k}+1}$, what is the smallest prime factor of $p$?

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