# 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

• Can you write a general formula for the sequence? Commented Apr 3, 2016 at 20:04
• What is the pattern here? $\qquad$ Commented Apr 3, 2016 at 20:06
• $a_{2n}=\frac {1}{n+1}$ and $a_{2n+1}=\frac {1}{n+3}$ Commented Apr 3, 2016 at 20:36
• @MichaelHardy 1; 3,2; 4,3; 5,4; …; n,n-1; n+1,n; … Commented Apr 3, 2016 at 20:36
• These type of questions make no-sense to me. If H(n) is a polynomial that vanishes at 0,1,2,3 then $a_{2n} = 1/(n+1) + H(n)$ and $a_{2n+1} = 1/(n+3) + H(n)$ also satisfies the given pattern yet $\{a_n\}$ does not converge to 0. Commented Nov 30, 2020 at 7:10

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.