Checking if series converge: $\sum_{k=1}^\infty \frac{(-1)^k}{\sqrt{k}}$ and $\sum_{k=1}^\infty \frac{(1+i)^k}{k!}$ etc. I want to see if these series converge:
$$\text{1.  }\sum_{k=1}^\infty \frac{(-1)^k}{\sqrt{k}}$$
$$\text{2.  }\sum_{k=1}^\infty \frac{(1+i)^k}{k!}$$
$$\text{3.  }\sum_{k=1}^\infty \frac{k^2+2}{k^4+1}$$
$$\text{4.  }\sum_{k=1}^\infty \frac{k^2 +2}{k^3+1}$$
Regarding $1.$ I used the alternating series test and got $a_k = \frac{1}{\sqrt{k}}$. The limit is $\lim_{k \to \infty} a_k = \frac{1}{\sqrt{k}} = 0$. So I have to show that $$a_k \geq a_{k+1}$$ $$\Leftrightarrow \frac{1\sqrt{k+1}}{\sqrt{k}} \geq 1$$ $$\Leftrightarrow \frac{\sqrt{k+1}}{\sqrt{k}} \geq 1$$ which is true.
Regarding $2.$ I used the ratio test which would give 
$$|\frac{a_k+1}{k!}| = \frac{\frac{(1+i)^{k+1}}{(k+1)!}}{\frac{(1+i)^k}{k!}} = \frac{(1+i)^{k+1}\cdot k!}{(k+1)! \cdot (1+i)^k} = \frac{(1+i)^{k+1} \cdot k!}{k! (k+1) \cdot (1+i)^k} = \frac{(1+i)^{k+1}}{(k+1)\cdot (1+i)^k}$$, but where do I go from here?
Regarding $3.$ I used the ratio test as well and got 
$$\lim_{k \to \infty} |\frac{a_{k+1}}{a_k}| = \frac{\frac{(k+1)^2+2}{(k+1)^4+1}}{\frac{(k^2+2)}{(k+1)^4+1}} = \frac{((k+1)^2+2) \cdot ((k+1)^4+1)}{((k+1)^4+1) \cdot (k^2 +2)} = \frac{(k+1)^2+2}{k^2+2} = \frac{k^2+2k+3}{k^2+2} = \frac{k^2 \cdot (\frac{2k}{k} + \frac{3}{k^2})}{k^2 \cdot \frac{2}{k}} = 2$$
Regarding $4.$ I just rewrote it as 
$$\lim_{k \to \infty} \frac{k^2 \cdot \frac{2}{k^2}}{k^3 \cdot \frac{1}{k^3}} = 0$$
Can someone please help me? (I just started learning about sequences)
 A: *

*What you did is fine.

*The absolute value of a complex number is always real and non-negative. So you have 
$$
\left|\frac{a_{k+1}}{a_k}\right|= \left|\frac{(1+i)^{k+1}}{(k+1)(1+i)^k}  \right|
=\left|\frac{(1+i)}{k+1}\right|=\frac{|1+i|}{k+1}=\frac{\sqrt2}{k+1}.
$$

*Your limit is wrong. It's $1$, which gives you no information. Here you would usually do comparison. Something like 
\begin{align}
\sum_{n=1}^\infty\frac{k^2+2}{k^4+1}
&\leq\sum_{n=1}^\infty\frac{k^2+2}{k^4}
=\sum_{n=1}^\infty\frac{1}{k^2}+\sum_{n=1}^\infty\frac{2}{k^4}<\infty.
\end{align}

*Here you also want to do comparison. The terms are basically $1/k$, so you want should expect this to diverge. You can do
$$
\sum_{n=1}^\infty\frac{k^2+2}{k^3+1}\geq\sum_{n=1}^\infty\frac{k^2}{k^2+k^3}=\frac12\,\sum_{n=1}^\infty\frac1k=\infty.
$$
===========================
If you are weak with inequalities, you can do the comparison in 3 and 4 by taking limits instead. Namely, if the number $a_n$ and $b_n$ are positive and $\lim_{n\to\infty}\tfrac{a_n}{b_n}$ exists and is not zero, then $\sum_na_n$ converges if and only if $\sum_nb_n$ converges. So in 3,
$$
\frac{\frac{k^2+2}{k^4+1}}{\frac1{k^2}}=\frac{k^2(k^2+2)}{k^4+1}=\frac{1+2/k^2}{1+1/k^4}\to1
$$
tells you that the series behaves like $\sum_k\frac1{k^2}$. And in 4 you can do 
$$
\frac{\frac{k^2+2}{k^3+1}}{\frac1k}=\frac{k(k^2+2)}{k^3+1}=\frac{1+2/k^2}{1+1/k^3}\to1.
$$
