Prove $a_n = (-1)^n\frac{n+1}{n}$ diverges Consider the sequence $\displaystyle \{a_n\}_{n=1}^{\infty}$ where
$a_n = \displaystyle (-1)^n\frac{n+1}{n}$. Prove that the sequence diverges. That is, prove that, for every $L\in{\rm I\!R}$, the
limit  of the sequence is not equal to $L$.
I need to come up with a formal proof to this problem, and don't know where to start.
 A: Have in mind that if such limit $L$ exists then every subsequence of $a_n$, say $b_n$ and $d_n$, must also converge to $L$.
But you can easily see in the expression that $a_n$ depends on $(-1)^n$ to be positive or negative. In fact, if $n = 2k$ then $(-1)^{2k} = 1$ and $a_n$ is positive. If $n = 2k + 1$ then $(-1)^{2k+1} = -1$ and $a_n$ is negative.
Now we can use this reasoning to write our proof.
Let $b_k = 2k$ and $d_k = 2k + 1$ with $k \in \mathbb{N}$. For $L$ to exist the following condition must be true:
$$L = \lim a_n \Rightarrow L = \lim a_{b_k} = \lim a_{d_k} $$
But
$$\lim a_{b_k} = (-1)^{2k}\frac{2k+1}{2k} = 1$$
$$\lim a_{d_k} = (-1)^{2k+1}\frac{2k+2}{2k+1} = -1$$
Therefore
$$\lim a_{b_k} \neq \lim a_{d_k}$$
And $a_n$ has no limit equal to $L$.
A: Since the values alternate between positive and negative the only possible limit is 0. Since the limit is not 0 (can you prove it?) then no limit exists.
A: Hint: If the limit exists then any subset of the series should converge to the same limit. Select n =2k (even). What is the limit for $k \to \infty$? Then select n = 2k+1. What is its limit for $k \to \infty$? If the limits are not the same, you can conclude that the limit does not exist.
