Nature of the series $\sum 1+(-1)^{n+1} (2n+1)$ .

Question

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?

Answer 1

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.

Answer 2

For a series to converge the general term must tend to $0$. In this case it doesn't so the series is divergent.

Answer 3

Surely oscillates infinitely since $$a_{n}=2n+2\quad,\quad n\text{ is odd}\\a_{n}=-2n\quad,\quad n\text{ is even}$$

Answer 4

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.