The question is as follows: Determine if the sequence:
$$\{1/1,\ 1/3,\ 1/2,\ 1/4,\ 1/3,\ 1/5,\ 1/4,\ 1/6,\ \ldots\}$$
diverges or converges. If it is convergent, find the limit.
I am not sure where to begin, I'd appreciate it if anybody could help me on this one.
Thankyou
As noted above, for all $n \in \mathbb N$:
$$a_{2n}=\frac1{n+1} \qquad \text{and}\qquad a_{2n+1}=\frac1{n+3}.$$
Or equivalently:
$$a_n = \left. \begin{cases} \frac 1 {1+n/2} = \frac 2 {n+2}, &\text{if } n \text{ even}, \\ \frac 1 {3+\frac{n-1}2} = \frac 2 {n+4}, &\text{if } n \text{ odd} \\\end{cases} \right\}\leq \frac2n.$$
(We just cut off the extra bit on the denominator.)
I.e., for all $n \in \mathbb N$, $$ 0 \le a_n \le \frac 2 n$$ and the squeeze theorem finishes it.
Starting with $1/3$, each term appears just twice.
Given a small positive number $\varepsilon$, make an integer $n$ so big that $1/n<\varepsilon$. Then move along the sequence until the second time that $1/n$ appears. All terms beyond that will be between $\varepsilon$ and $0$. Since this can be done no matter how small $\varepsilon$ gets, the sequence converges to $0$.
$a_{2n}=\frac {1}{n+1}$ and $a_{2n+1}=\frac {1}{n+3}$ So $\lim_{n \to \infty} a_{2n}=0$ and $\lim_{n \to \infty} a_{2n+1}=0$. So $\lim_{n \to \infty} a_{n}=0$. So the sequence is convergent and its limit is $0$.
For a given $\epsilon$, let $k=\frac{2}{\epsilon}$. Then for all $n>2k,\ x_n<\frac{1}{k}<\epsilon$.
The approach of Bérénice is nice and looks quite reasonable, since the first step is to identify the elements of the sequence \begin{align*} \begin{array}{ccccccccc} a_0&\ a_1&\ a_2&\ a_3&\ a_4&\ a_5&\ a_6&\ a_7&\ \cdots\\ \color{blue}{1}&\ \frac{1}{3}&\ \color{blue}{\frac{1}{2}}&\ \frac{1}{4}&\ \color{blue}{\frac{1}{3}}&\ \frac{1}{5}&\ \color{blue}{\frac{1}{4}}&\ \frac{1}{6}&\ \cdots\\ \\ \end{array} \end{align*} which typically results after some analysis in a separation of even and odd elements: \begin{align*} a_{2n}=\frac{1}{n+1}\qquad\qquad a_{2n+1}=\frac{1}{n+3}\qquad\qquad n\geq 0\tag{1} \end{align*}
With (1) in mind we can continue along different lines. One of them is to see the elements are positive and essentially growing as $\frac{1}{n}$, So, we could think of squeezing the sequence between the constant sequence $(0)_{n\geq 0}$ and a sequence related with $\left(\frac{1}{n+1}\right)_{n\geq 0}$ where we use $n+1$ in the denominator to avoid division by zero when $n=0$.
Starting from (1) we obtain \begin{align*} \color{blue}{a_n}&=\frac{1}{\frac{n}{2}+1}\left(\frac{1+(-1)^n}{2}\right)+\frac{1}{\frac{n-1}{2}+3}\left(\frac{1-(-1)^n}{2}\right)\\ &=\frac{1}{n+2}\left(1+(-1)^n\right)+\frac{1}{n+5}\left(1-(-1)^n\right)\\ &\leq \frac{2}{n+2}+\frac{2}{n+5}\\ &\,\,\color{blue}{\leq \frac{4}{n+2}} \end{align*}
We derive this way an upper bound $\frac{4}{n+2}$ for $a_n$ and can conclude from \begin{align*} 0\leq a_n\leq \frac{4}{n+2}\quad\longrightarrow\quad 0 \end{align*} the sequence $(a_n)_{n\geq 0}$ is a zero-sequence.
The sequence $1, 1, \frac 12, \frac 12, \frac 13, \frac 13, \dots$ is a sequence that is greater than or equal to yours and converges to zero. The sequence $0, 0, 0, \dots$ is less than or equal to it and converges to zero, so by the squeeze theorem your sequence also converges to zero.
