<s>[INST] 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
 [/INST] 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.
</s>