Proving the limit of an alternating sequence

Question

Prove that $S_n$ converges to $0$ as $n\rightarrow$ $\infty$, where

$$ S_n = \left\{ \begin{array}{cc} 2/n & \text{ if $n$ is odd} \\ n/(n^2-1) & \text{ if $n$ is even }\end{array}\right. $$

I want to check my proof as I feel that I am leaving something out.

Proof:

Fix $\epsilon >0$. Since $\frac{2}{n} \rightarrow 0$, $\exists N_1$ such that $|\frac{2}{n} -0|=\frac{2}{n}<\epsilon$. Likewise, since $\frac{n}{n^2-1} \rightarrow 0$, $\exists N_2$ such that $|\frac{n}{n^2-1}-0|=\frac{n}{n^2-1} < \frac{1}{n-3} <\epsilon,\forall n \in \mathbb{N}$ (since $3n-1 > 0, \ \forall n \in \mathbb{N}$).

Now choose $N_1 =\frac{2}{\epsilon}$, and $N_2 = \frac{1+3\epsilon}{\epsilon}$. Let $M = max\{N_1,N_2\}$, then $\forall n>M$:

If n is odd, then $|S_n - 0| = |\frac{2}{n} -0|=\frac{2}{n} < \frac{2}{M}<\epsilon$. If n is even, then $|S_n - 0| = |\frac{n}{n^2-1}-0|=\frac{n}{n^2-1} < \frac{1}{n-3} < \frac{1}{M-3} <\epsilon$. In both cases, $|S_n - 0|<\epsilon$. Done.

Is the proof complete? My concern is with choosing $M$ to be max. Usually, with simple sequences I can choose $N$ to be a single value (e.g. $2/\epsilon$)and plug it in $2/N$ to get $\epsilon$ but it seems weird to have the max of two numbers and the result is $\epsilon$.

Answer 1

The proof is correct, more generally $(u_n)$ converges if and and only if $(u_{2n})$ and $(u_{2n+1})$ converge toward the same limit.

Answer 2

Hint

I think the following will better help.$$0\le S_{n+1}\le {2\over n}$$

Answer 3

Since $$ \frac{n} {{n^2 - 1}} \leqslant \frac{n} {{n^2 - \frac{{n^2 }} {2}}} = \frac{n} {{\frac{{n^2 }} {2}}} = \frac{2} {n} $$ you have that for every $n \geq 2$ it is $S_n \leq \frac{2}{n}$. Therefore it is enough to prove that $$ \forall \varepsilon > 0\,\,\,\exists n\left( \varepsilon \right) \in \mathbb{N}:\,\,\,\,n > n\left( \varepsilon \right) \Rightarrow \left| {\frac{2} {n}} \right| < \varepsilon $$ in order to obtain that $$ \forall \varepsilon > 0\,\,\,\exists n\left( \varepsilon \right) \in \mathbb{N}:\,\,\,\,n > n\left( \varepsilon \right) \Rightarrow \left| {S_n } \right| < \varepsilon $$ If $$ \left| {\frac{2} {n}} \right| < \varepsilon $$ you get that $n> \frac{2}{\varepsilon}=n(\varepsilon)$.