Elementary proof that $\sum_{n \geq 1} \frac{(-1)^n}{\sqrt{n}}+i \frac{1}{n^2}$ is conditionally convergent. I must prove that the complex series $$\sum_{n \geq 1} \frac{(-1)^n}{\sqrt{n}}+i \frac{1}{n^2}$$ is conditionally convergent. The catch is, I'm not supposed to use anything "too advanced", meaning I only have in hands the ratio test and the root test, which are pretty much useless, since the failure of the ratio test implies the failure of the root test. We can use the comparison test too.
I managed to prove it using Raabe's test, here goes:
$$\sum_{n \geq 1} \left| \frac{(-1)^n}{\sqrt{n}}+i \frac{1}{n^2}\right| = \sum_{n \geq 1} \sqrt{\frac{1}{n}+\frac{1}{n^4}},$$ the ratio test and root test both fail for $\sqrt{\frac{1}{n}+\frac{1}{n^4}} = \frac{\sqrt{n^3+1}}{n^2}$, but by Raabe's test: $$\lim_{n \to +\infty} n\left(1 - \frac{n^2}{(n+1)^2}\sqrt{\frac{(n+1)^3+1}{n^3+1}}\right) = \frac{1}{2} < 1,$$ the series diverges (actually, I used Wolfram Alpha to get this limit, cheated again).
Now, we know that the complex series converges if and only if $\sum_{n \geq 1}\frac{(-1)^n}{\sqrt{n}}$ and $\sum_{n \geq 1}\frac{1}{n^2}$ both converge. The second one obviously converge. And for the first one, both the ratio test and the root test fail, but by Raabe's test: $$\lim_{n \to +\infty} n\left(1+\sqrt{\frac{n}{n+1}}\right) = +\infty > 1,$$and the series converges.

Question: Is there an elementary approach to this?

 A: You just need to break it up into real and imaginary parts. Consider
$$\sum_{n=1}^\infty (a_n+b_ni)$$
with $a_n,b_n\in\Bbb R$. Then given $\epsilon >0$ if $\sum a_n=a, \sum b_n=b$ we have some $N'(\epsilon),N''(\epsilon)$ so that when $k>N', m>N''$ we have
$$\left|\sum_{n=1}^k a_n-a\right|<\epsilon, \left|\sum_{n=1}^{m} b_n-b\right|<\epsilon$$
but then if $\ell>N=\max\{N',N'' \}$ we have
$$\left|\sum_{n=1}^\ell (a_n+b_ni)-a+bi\right|<\left|\sum_{n=1}^\ell a_n-a\right|+\left|i\left(\sum_{n=1}^\ell b_n-b\right)\right|<2\epsilon.$$
But then both of your original series are convergent, the real part series by the alternating series test since $n^{-1/2}$ is a monotone decreasing function, and the imaginary part is absolutely convergent.
A: Consider the sequence of partial sums $S_n=\sum_{k=1}^n \left[\frac{(-1)^k}{\sqrt{k}} + i\frac{1}{k^2}\right]$. We have
$$ S_{n} = S_{n-1} + \frac{(-1)^{n}}{\sqrt{n}} + i\frac{1}{n^2} $$
From earlier work, we see
$$ \left|S_n-S_{n-1}\right|=\sqrt{\frac{1}{n}+\frac{1}{n^4}}. $$
Thus, $|S_n-S_{n-1}|$ monotonically approaches 0, implying $S_n \to S$. The original series must converge. [edit]This is very wrong. See comments.[/edit] However, it is conditional, since
$$ \sum_{n=1}^\infty \left|\frac{(-1)^n}{\sqrt{n}}+i\frac{1}{n^2}\right| = \sum_{n=1}^\infty\sqrt{\frac{1}{n}+\frac{1}{n^4}} > \sum_{n=1}^\infty \frac{1}{\sqrt{n}} \to \infty. $$
A: Since $$\sqrt{\frac{1}{n} + \frac{1}{n^4}} \ge \sqrt{\frac{1}{n}} = \frac{1}{\sqrt{n}}$$ and $\sum_{n = 1}^\infty \frac{1}{\sqrt{n}}$ diverges, by direct comparison the series 
$$\sum_{n = 1}^\infty \left|\frac{(-1)^n}{\sqrt{n}} + \frac{i}{n^2}\right| = \sum_{n = 1}^\infty\sqrt{\frac{1}{n} + \frac{1}{n^4}}$$ diverges.
To see $\sum_{n = 1}^\infty (-1)^n/\sqrt{n}$ converges, let $s_n$ be the sequence of partial sums of the series $\sum_{n = 1}^\infty (-1)^{n-1}/\sqrt{n}$. Then 
$$s_{2n} = s_{2n-2} + \frac{1}{\sqrt{2n-1}} - \frac{1}{\sqrt{2n}} \ge s_{2n-2}$$
and
$$s_{2n} = 1 - \left(\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{3}}\right) - \cdots - \left(\frac{1}{\sqrt{2n-2}} - \frac{1}{\sqrt{2n-1}}\right) - \frac{1}{\sqrt{2n}} \le 1.$$
for all $n\in \Bbb N$. So the sequence $\{s_{2n}\}$ is increasing and bounded above by $1$. By the monotone convergence theorem, $s_{2n}$ converges. Let $s = \lim_{n\to \infty} s_{2n}$. Then 
$$\lim_{n\to \infty} s_{2n+1} = \lim_{n\to \infty} \left(s_{2n} + \frac{1}{\sqrt{2n+1}}\right) = \lim_{n\to \infty} s_{2n} = s.$$
Since $\lim_{n\to \infty} s_{2n} = s = \lim_{n\to \infty} s_{2n+1}$, $\{s_n\}$ converges (with limit $s$). So $\sum_{n = 1}^\infty (-1)^n/\sqrt{n}$ converges (to $-s$).
