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?