Find the general term of the sequence $a_n = (0, 3, 1/2 , 5/3 , 2/3 , 7/5 , 3/4 , ...)$. Then find the limit of this sequence.

Question

I have an exercise in which I am stuck. The exercise is

Find the general term of the sequence $$a_n = (0, 3, \dfrac{1}{2}, \dfrac{5}{3}, \dfrac{2}{3}, \dfrac{7}{5}, \dfrac{3}{4}, ...)$$. Then find the limit of this sequence.

So far, I have tried like this(Idk if its right)

First thing, I have written sequence as like this

$$a_n = (0, 3, \dfrac{1}{2}, \dfrac{5}{3}, \dfrac{2}{3}, \dfrac{7}{5}, \dfrac{3}{4}, ...)$$

$$indexes \rightarrow 1, 2, 3,\hspace{0.5mm} 4,\hspace{1mm}5, 6, \hspace{0.5mm}7, 8, \hspace{2.2cm}$$

Now, I can get/ divide indexes in odd and even.

For Odd indexes: $$a_n = (0, \dfrac{1}{2}, \dfrac{2}{3}, \dfrac{3}{4}, ...)$$ $$indexes \rightarrow \hspace{4mm}1,\hspace{2mm} 2, \hspace{1mm}3,\hspace{2.5mm} 4,..\hspace{2cm}$$

For Even indexes: $$a_n = (3, \dfrac{5}{3}, \dfrac{7}{5}, ...)$$ $$indexes \rightarrow \hspace{4mm}1,\hspace{2mm} 2, \hspace{1mm}3,..\hspace{2cm}$$

So Now, I know the sequence of odd and even indexes(I think):

$$Odd: \dfrac{n}{n+1}\ and\ even\ \dfrac{n+2}{n}\ \forall n : n=2(k+1)$$

I have arrived here. I don't know how to proceed. Thanks For all help.

Answer 1

Taking the first term as $\frac 01$, the second as $\frac 31$ and the pattern continues as given, then the general term is $$a_n=\frac{n+(-1)^n}{n-(-1)^n}$$

Clearly the limit, as $n\rightarrow\infty=1$

Answer 2

$$a_{2n+1}=\frac {2n+3}{2n+1} $$

$$a_{2n}=\frac {n}{n+1} $$

$$\lim a_{2n}=1=\lim a_{2n+1} \implies $$ $$\lim a_n=1$$

Answer 3

User Fatima gives $a_{2n}$ and $a_{2n+1}$, i.e the even and odd terms of the sequence $a_n$.

These $2$ subsequences, $a_{2n}$, $a_{2n+1}$ of the sequence $a_n$ converge to the same limit L.

Remains to be shown that $a_n$ converges .

Revised appendix:

Let $\epsilon >0$ be given.

1)$a_{2n}$ is convergent,

there is a $n_1$ such that for $n \ge n_1$ :

$ |a_{2n} - L| \lt \epsilon$.

2)$a_{2n+1}$ converges,

there is a $n_2$ such that for $n \ge n_2$:

$|a_{2n+1}- L| \lt \epsilon$.

3) Set $n_3 = \max(2n_1,2n_2+1)$, then

for $n \gt n_3$ :

$|a_n- L| \lt \epsilon$,

hence convergent.