How to find if a sequence is converging or diverging? 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

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.
