Is it not divergent?

For a given integer $m=10^3+1$ , we can upper-bound the tail of the series $\sum_{n=m}^{\infty}(2^{1/n^2}-2^{1/n^3})$ as follows $\sum_{n=m}^{\infty}(2^{1/n^2}-2^{1/n^3}) \leq \sum_{n=m}^{\infty}(2^{1/n^2}-1) = \sum_{n=m}^{\infty} (\exp(\ln(2)/n^2)-1) \stackrel{(a)} {\leq} \sum_{n=m}^{\infty}2\ln(2)/n^2 \stackrel{(b)}{\leq} 2 \ln(2)/10^3 \leq 0.0014$ Where in (a), we have used the inequality that $\exp(x)\leq 1+2x, 0\leq x\leq 1$ and the summation in (b) has been upper-bounded by the corresponding integral. Thus it follows that to evaluate $\lfloor 100S\rfloor$ , it is enough to evaluate the series up to $m=10^3$ and then take the floor. This finite summation has been evaluated by a computer program.

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#include<iostream>
#include<cmath>
using namespace std;

int main()

{
    int i=1;
    float s=0;
    do{
        s=s + pow(2,pow(i,-2)) - pow(2,pow(i,-3));
        i++;
    }
    while(i<=10000);
    cout<<floor(100*s);
    return 0;
}

When $i$ was set till $100$ , it returned $31$ .

When $i$ was set till $1000$ , it returned $32$ .

When $i$ was set till $10000$ , it again returned $32$ .

So it is apparent that the sum has converged enough till then. So, $\boxed{32}$ is the final answer.