I would like to show that:
$ \sum_{k=n+1}^{\infty} e^{-k^2} \sim e^{-(n+1)^2}$
We have:
$ \forall p\geq2$
$ \exp((n+1)^2-(n+p)^2)=\exp(2n(1-p)+1-p^2)=o(1)$
$ e^{-(n+p)^2}=o \left( e^{-(n+1)^2} \right)$
So:
$ \sum_{k=n+1}^{\infty} e^{-k^2} =e^{-(n+1)^2}+o \left( e^{-(n+1)^2} \right)\sim e^{-(n+1)^2}$
Given that there are infinitely many $ o \left(e^{-(n+1)^2} \right)$, can the equality $ \sum_{k=n+1}^{\infty} e^{-k^2} =e^{-(n+1)^2}+o \left( e^{-(n+1)^2} \right) $ be directly written?