Does $\sum_{k=0}^\infty e^{-\sqrt{k}} (-1)^k$ converge faster than $e^{-\sqrt{k}}$? Does $\sum_{k=0}^\infty e^{-\sqrt{k}} (-1)^k$ converge faster than $e^{-\sqrt{k}}$? In particular, is
$$
\lim_{N\to\infty} e^{\sqrt{N}} \sum_{k=N}^\infty e^{-\sqrt{k}} (-1)^k=0?
$$
I know that when the $\sqrt{k}$ and $\sqrt{N}$ are replaced with $k$ and $N$ the expression in the limit oscillates around $0$ alongside the parity of $N$. The square root effectively makes the sign of that summand wiggle faster, so I suspect that the extra cancellation makes the series converge faster.
Any help would be appreciated. Thanks!

$$e^{\sqrt{N}}\sum_{k\geq N}e^{-\sqrt{k}}(-1)^k = (-1)^N \sum_{h\geq 0}(-1)^h \exp\left(-\frac{h}{\sqrt{N+h}+\sqrt{N}}\right) $$
is well-approximated by
$$ (-1)^N\sum_{h\geq 0}(-1)^h \exp\left(-\frac{h}{2\sqrt{N}}\right) = \frac{(-1)^N}{1+\exp\left(-\frac{1}{2\sqrt{N}}\right)}$$
hence the given limit does not exist.
