convergence of $\sum\limits_{n=1}^\infty \frac{(-1)^{\lfloor \sqrt{n}\rfloor}}{\sqrt{n}}$ Does anybody know how to show analytically the convergence (divergence ?) of $\enspace\displaystyle \sum\limits_{n=1}^\infty \frac{(-1)^{\lfloor \sqrt{n}\rfloor}}{\sqrt{n}}\enspace$ or some literature for it ? (Thanks!) 
Note: Unfortunately I am still missing a useful approach. 
EDIT:
I want to thank all here for the nice help. 
Result: The series is divergent (but limited) as it is written. 
It's possible to change to $\enspace\displaystyle\sum\limits_{n=1}^\infty (-1)^n (-2+\sum\limits_{k=n^2}^{(n+1)^2-1}\frac{1}{\sqrt{k}})\enspace$ to avoid the oscillation and to get convergence.    
 A: You can prove that it diverges by observing that the sum of each "block" is greater than a constant, that is, $$\sum_{k=n^2}^{(n+1)^2-1}\frac{1}{\sqrt{k}}>C$$for some positive $C$.
A: Sorry, I have forgotten how to type in LateX, so here it is by hand.
Basically, this is an explicit version of the other answer. 
Essentially, we show that each block has a minimal magnitude (infimum) by which it add or subtracts from the partial sum. This non-decreasing (constant) bound makes the series fail the criteria for the alternating series test (I believe)


Interestingly, if we study the blocks of partial terms, they tend towards "$2$" in magnitude, so the series diverges more "hectically" as we take $n$ to infinity than our bound may suggest.  
A: This does not converge;
it oscillates by
approximately $2$.
If
$m^2 \le m
\le m^2+2n
$,
then
$\lfloor \sqrt{n}\rfloor
= m$.
Therefore
$\begin{array}\\
S(M)
&=\sum_{n=1}^{M^2+2M} \frac{(-1)^{\lfloor \sqrt{n}\rfloor}}{\sqrt{n}}\\
&=\sum_{m=1}^{M}
\sum_{n=m^2}^{m^2+2m} \frac{(-1)^m}{\sqrt{n}}\\
&=\sum_{m=1}^{M}
(-1)^m\sum_{n=m^2}^{m^2+2m} \frac{1}{\sqrt{n}}\\
&=\sum_{m=1}^{M}
(-1)^ms(m)
\qquad\text{where }s(m)=\sum_{n=m^2}^{m^2+2m} \frac{1}{\sqrt{n}}\\
\end{array}
$
$s(m)
\lt \frac{2m+1}{m}
= 2+\frac1{m}
$
and
$s(m)
\gt \frac{2m+1}{m+1}
=2-\frac1{m+1}
$.
Therefore,
as $n$ goes from
$m^2$ to
$m^2+2n$,
the value of the sum
changes by approximately $2$,
increasing when $n$ is even
and
decreasing when $n$ is odd.
Therefore the series does not converge,
since it oscillates.
However,
there might be a sense
in which the series converges
since we can write
$S(2M)
=\sum_{m=1}^{2M}(-1)^ms(m)
=\sum_{m=1}^{M}(s(2m)-s(2m-1))
$.
I can show that that
$s(m)-s(m+1)
\to 0$,
but I can not yet show that
$\lim_{M \to \infty}\sum_{m=1}^{M}(s(2m)-s(2m-1))
$
exists.
Here's what I have so far,
and I will leave it at this:
$\begin{array}\\
s(m)-s(m+1)
&\gt (2-\frac1{m+1})-(2+\frac1{m+1})\\
&= -\frac{2}{m+1}\\
\text{and}\\
s(m)-s(m+1)
&\lt (2+\frac1{m})-(2-\frac1{m+2})\\
&= \frac1{m}+\frac1{m+2}\\
&= \frac{2m+2}{m(m+2)}\\
\text{so}\\
s(m)-s(m-1)
&= O(\frac1{m})\\
\end{array}
$
A: It's alternating series (with periodically strange alternation, but alternation nonetheless) and thus converges according to alternating series test.
