Is that last term $2ne^{n^2}$ or is it $2^ne^{n^2}?$ I'm going to assume the latter, unless told otherwise. Under that assumption: Let $b_n=1+2e+4e^4+8e^9+\cdots+2^ne^{n^2}$ $=2^ne^{n^2}\left(1+\frac12e^{-2n+1}+\frac14e^{-4n+4}+\cdots+\frac1{2^n} e^{-n^2}\right)$ $<2^ne^{n^2}\left(1+\frac12+\frac14+\cdots\right)<2\cdot2^ne^{n^2}$ So, $2^ne^{n^2}<b_n<2\cdot2^ne^{n^2}$ Take logarithms: $n^2+n\ln2<a_n<n^2+(n+1)\ln2.$ $1+\frac1n\ln 2<\frac{a_n}{n^2}<1+\frac{n+1}{n^2}\ln2.$ So the limit is $1.$

Actually, the last term is $2ne^{n^2}$. Nevertheless, it can be done in a similar way. By the way, the limit is also $1$.