Is this Series Convergent and does it also Absolutely Converge? $$\sum\limits_{n=1}^{\infty}a_{n}\ \text{where}\ a_{n} =
\left\{
 \begin{array}{ll}
  \frac{1}{n}  &  n\mbox{ is even} \\
  \frac{-1}{n^2}  &  n\mbox{ is not even}
 \end{array}
\right.$$
I need to check if this series converges and also converges absolutely... I was thinking about looking at the elements and see a pattern but I  don't see any patterns... any directions please?
 A: Claim: This series does not converge.
Proof: 
Assume by means of contradiction that the above series converges (credit to @WillR for noting we need to begin with this). Then the above series is equal to:
$$\sum\limits_{n=1}^{\infty} \frac{1}{2n} + \sum\limits_{n=0}^{\infty} -\frac{1}{(2n+1)^2}$$
(To see this, just check that it works for partial sums and take the limit; the expression is meaningful if the original series is absolutely convergent, if the original series is only conditionally convergent it does not make sense; however that case is covered below.)
We know that the harmonic series $\sum\limits_{n=1}^{\infty} \frac{1}{n}$ diverges. For a proof of this fact, see https://web.williams.edu/Mathematics/lg5/harmonic.pdf -- it is good to memorize this result.
Assume by means of contradiction then that $\sum\limits_{n=1}^{\infty} \frac{1}{2n}$ converged, say to a value $c$.
Then we would have:
$$c= \sum\limits_{n=1}^{\infty} \frac{1}{2n} = \sum\limits_{n=1}^{\infty} \frac{1}{2} \cdot \frac{1}{n} = \frac{1}{2}\sum\limits_{n=1}^{\infty} \frac{1}{n}$$
which would imply that the harmonic series converged, which is a contradiction.
The above is sufficient to show that your series is not absolutely convergent, but we want to rule out the possibility that it might be conditionally convergent too.
This follows from the fact that 
$$\sum\limits_{n=1}^{\infty} \frac{1}{n^2}$$
is absolutely convergent; i.e. this implies that the sum of the subsequence 
$$\sum\limits_{n\ \text{odd}} \frac{1}{n^2}$$ is also finite, which is the same number as 
$$-\left(\sum\limits_{n=0}^{\infty} -\frac{1}{(2n+1)^2}\right),$$
i.e. your series equals 
$$\infty - \sum\limits_{n\ \text{odd}} \frac{1}{n^2} = \infty$$
so it is not even conditionally convergent.
A: This series diverges. The subset of terms with odd index form an absolutely convergent series (remember that $\sum_n n^{-2}$ is absolutely convergent). The set of even index terms form a divergent series. In fact, $\sum_n n^{-1}$ is the divergent armonic series. Even if you retain only the even terms, the seies diverges (you should be able to prove it)
A: We write the partial sum
$$S_{2n+1}=\sum_{k=1}^{2n+1}a_k=\sum_{k=1}^na_{2k}+\sum_{k=1}^n a_{2k+1}=\frac12\sum_{k=1}^n\frac1k-\sum_{k=1}^n\frac1{(2k+1)^2}=\frac12H_n-v_n$$
where $(H_n)$ is a divergent sequence and $(v_n)$ is convergent. Hence $(S_{2n+1})$ is divergent and then the given series is also divergent.
A: Let $b_k$ be the sequence $b_k=\frac{1}{k}$ and $c_k$ be the sequence where $c_k=\frac{1}{(2k-1)^2}$.  Let $B_k$ be the $k$-th partial sum of the series of $b_k$'s and $C_k$ the $k$-th partial sum of the series of $c_k$'s.  Now, consider the partial sum of the given series $S_{2k}=\frac{1}{2}B_k-C_{k}$.  We know that $C_k$ converges as the $p$-series with $p=2$ converges and $B_k$ diverges as it is a subsequence of a harmonic series.  Therefore,
$$
\lim_{k\rightarrow\infty}S_{2k}=\frac{1}{2}\lim_{k\rightarrow\infty}B_k-\lim_{k\rightarrow\infty}C_k.
$$
The first limit is $\infty$ while the second limit is finite.  Therefore, the given series diverges.
