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
 A: 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.
A: 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: 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.
A: For a given $\epsilon$, let $k=\frac{2}{\epsilon}$. Then for all $n>2k,\ x_n<\frac{1}{k}<\epsilon$. 
A: $a_{2n}=\frac {1}{n+1}$ 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 $0$.
A: 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.
