Nature of the series $\sum 1+(-1)^{n+1} (2n+1)$ . The series $\sum 1+(-1)^{n+1} (2n+1)$ .is 
1. Convergent 
2. Oscillates finitely
3. Divergent
4. Oscillates infinitely
I found first few terms of this series, which are 4-4+8-8+...
So it seems like I will get such pairs if I expand the series more. But what can we conclude about the nature of the series at infinity? The series is oscillating infinitely. So can I say it is divergent?
 A: If $a_n=1+(-1)^{n+1}(2n+1)$, the $(a_n)$ is unbounded, hence $(a_n)$ does not converge to $0$. Thus $\sum a_n $ is divergent.
A: For a series to converge the general term must tend to $0$. In this case it doesn't so the series is divergent. 
A: Surely oscillates infinitely since $$a_{n}=2n+2\quad,\quad n\text{ is odd}\\a_{n}=-2n\quad,\quad n\text{ is even}$$

A: Suppose for contradiction that  the series $\sum 1+(-1)^{n+1} (2n+1)$ converges, then the sequence $$1+(-1)^{n+1} (2n+1)\to 0,\;\;\text{ as }n\to\infty.$$
However, \begin{align} c_n:= 1+(-1)^{n+1} (2n+1)=\begin{cases}2n+2,&\text{if}\;n\;\text{is even,}\\-2n,&\text{if}\;n\;\text{is odd.}\end{cases}  \end{align}
\begin{align}\text{Thus, for odd} \; n\in\Bbb{N},\; c_n= -2n\to -\infty,\;\text{while}\; c_n= 2n+2\to \infty\;\text{for even} \;n,\text{contradiction.}\end{align}
Therefore, option $(1)$ is eliminated. Now, by Direct Comparison
\begin{align}\infty= \sum^{\infty}_{n=1} n\leq\sum^{\infty}_{n=1} (2n+2)\leq \infty\;\text{and} \;-\infty\leq\sum^{\infty}_{n=1} (-2n)\leq\sum^{\infty}_{n=1} -n= -\infty\end{align}
Option $(3)$ is eliminated because 
\begin{align}\sum^{\infty}_{n=1} 1+(-1)^{n+1} (2n+1)=\infty\; \text{if}\;n\;\text{is even}\end{align}
\begin{align}\text{and}\;\sum^{\infty}_{n=1} 1+(-1)^{n+1} (2n+1)=-\infty\; \text{if}\;n\;\text{is odd.}\end{align}
Hence, it oscillates infinitely.
