Epsilon-Delta Proof of Divergence of $S_n = (-1)^n*n$

Question

I am self-learning Analysis (reading Spivak's Calculus) but I found this problem in Ross' Elementary Analysis that I found interesting. However, I am having some difficulty proving the statement. It is: Show that the following sequence does not converge, $$ S_n = (-1)^n*n $$. I am attempting a proof via contradiction; ie: assume $ \exists $ a limit L $ \mid \forall \epsilon > 0, \exists $ N $ \in \mathbb{N} \mid n > $ N$ \implies |S_n - L| < \epsilon. \\ \implies |(-1)^nn - L| < \epsilon. $

Now i am stuck here because i am unsure which epsilon to choose and what to work towards in order to obtain my contradiction. Am I required to show that for all L $ \in \mathbb{R}$ that there is an $ \epsilon > 0 $ such that for any $ N > 0 $ there is an $ n > N $ such that $ |S_n - L| \ge \epsilon $?

Any help is appreciated.

Answer 1

Assume that $(S_n)$ converges. Then there exists $l\in \mathbb{R}$ such that $(S_n)$ converges to $l$. Then either $l\ne 0$ or $l=0$.

Case 1: Suppose $l\ne 0$. Then there exists $N\in \mathbb{N}$ such that for each $n>N$, $|(-1)^nn-l|<\dfrac{|l|}{2}$. Then we have $|n-|l||=||(-1)^nn|-|l||\le|(-1)^nn-l| <\dfrac{|l|}{2}\Rightarrow n<\dfrac{3|l|}{2}$ for each $n>N$. This is a contradiction.

Case 2: Suppose $l= 0$. Then there exists $N_1\in \mathbb{N}$ such that for each $n>N_1$, $|(-1)^nn|<1$. Hence there exists $N\in \mathbb{N_1}$ such that for each $n>N_1$, $n<1$. Contradiction.

Hence $(S_n)$ diverges.

Answer 2

Hints:

• Write down the elements of the sequence first see what is happening.
• You can choose any $\epsilon $ to get a contradiction. The convergence definition does not work for any $\epsilon$ with this sequence.
• Use the Archimedean property: "Given any real number $l$ (wink) there exists $n \in \Bbb N$ for which $l \lt n$"

As for you question on the bottom. Yes, that is exactly what you are supposed to prove.