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!
 A: $$e^{\sqrt{N}}\sum_{k\geq N}e^{-\sqrt{k}}(-1)^k = (-1)^N \sum_{h\geq 0}(-1)^h \exp\left(-\frac{h}{\sqrt{N+h}+\sqrt{N}}\right) $$
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.
A: Yes, sort of.  Let $S=\sum_{k=0}^\infty (-1)^ke^{-\sqrt{k}}$.  Then a basic error bound on partial sums of alternating series tells us
$$\left|S-\sum_{k=0}^N (-1)^ke^{-\sqrt{k}}\right|<\left|(-1)^{N+1}e^{-\sqrt{N+1}}\right|=e^{-\sqrt{N+1}}$$
And we know that $L=\lim_{k\rightarrow \infty} e^{-\sqrt{k}}=0 $, so the distance for $k=N$  to the limit is $\left|L-e^{-\sqrt{N}}\right|=e^{-\sqrt{N}}$.
So, since $e^{-\sqrt{N}}>e^{-\sqrt{N+1}}$, we have that $\sum_{k=0}^\infty (-1)^ke^{-\sqrt{k}}$ converges faster in that it is always closer to its limit.  However, often 'converges faster than' is meant about the ratio of their closeness to their limits going to 0, this doesn't support such a claim, though it could be true.
