No problem, happy to help! (Glad to hear it)

[mathmari]

Hey!

I want to check which of the following sequences converges and from those that don't converge I want to check if it has a convergent subsequence.

1. $\displaystyle{1, 1-\frac{1}{2}, 1, 1-\frac{1}{4}, 1, 1-\frac{1}{6}, \ldots}$
2. $\displaystyle{1, \frac{1}{2}, 1, \frac{1}{4}, 1, \frac{1}{6}, \ldots}$
3. $\displaystyle{c_n=\frac{2n+5}{3-n}, \ n\in \mathbb{N}}$
4. $\displaystyle{d_n=\frac{n^2+1}{n+1}, \ n\in \mathbb{N}}$

I have done the following:

1. $\displaystyle{1, 1-\frac{1}{2}, 1, 1-\frac{1}{4}, 1, 1-\frac{1}{6}, \ldots}$
   
   We have that
   \begin{equation*}a_n=\left\{\begin{matrix}
   1 & \text{ for odd } n \\
   1-\frac{1}{n} & \text{ for even } n
   \end{matrix}\right.\end{equation*} For odd $n$, $n=2k+1$, we have that $1$ converges to $1$, so the limit of $a_{2k+1}$ is $1$.
   
   For even $n$, $n=2k$, we have that $1-\frac{1}{n}$ converges to $1$, so the limit of $a_{2k}$ is $1$. The only limit point of the sequence $(a_n)_{n\in \mathbb{N}}$ is therefore $1$. We have that the sequence is bounded, since $1$ is a constant, $|1|\leq 1$ and $\displaystyle{0<\frac{1}{n}\leq 1 \Rightarrow -1\leq -\frac{1}{n}<0\Rightarrow 0\leq 1-\frac{1}{n}<1}$. Since the sequence is bounded and has one limit point, it is convergent and the limit is $1$.
2. $\displaystyle{1, \frac{1}{2}, 1, \frac{1}{4}, 1, \frac{1}{6}, \ldots}$
   
   We have that
   \begin{equation*}b_n=\left\{\begin{matrix}
   1 & \text{ for odd} n \\\
   \frac{1}{n} & \text{ for even } n
   \end{matrix}\right.\end{equation*}
   
   For odd $n$, $n=2k+1$, we have that $1$ converges to $1$,so the limit of $b_{2k+1}$ is $1$.
   
   For even $n$, $n=2k$, we have that $\frac{1}{n}$ converges to $0$, so the limit of $b_{2k}$ is $0$. The limit points of the sequence $(b_n)_{n\in \mathbb{N}}$ are therefore $1$ and $0$. So, the sequence $(b_n)_{n\in \mathbb{N}}$ diverges, since it approximates no number, because it is between $0$ and $1$. The divergent sequence has two convergent subsequences $b_{2k+1}, b_{2k}$.
   
   The subsequence $(b_{2k+1})_{k\in \mathbb{N}}$ is bounded and its only limit point is $1$. So, the subsequence $(b_{2k+1})_{k\in \mathbb{N}}$ converges $1$.
   
   The subsequence $(b_{2k})_{k\in \mathbb{N}}$ is bounded and its only limit point is $0$. So the subsequence $(b_{2k})_{k\in \mathbb{N}}$ converges to $0$.
3. $\displaystyle{c_n=\frac{2n+5}{3-n}, \ n\in \mathbb{N}}$
   
   We have that \begin{equation*}\lim_{n\rightarrow \infty}c_n=\lim_{n\rightarrow \infty}\frac{2n+5}{3-n}=\lim_{n\rightarrow \infty}\frac{n\left (2+\frac{5}{n}\right )}{n\left (\frac{3}{n}-1\right )}=\lim_{n\rightarrow \infty}\frac{2+\frac{5}{n}}{\frac{3}{n}-1 }=\frac{2+ 0}{0-1 }=-2\in \mathbb{R}\end{equation*}
   
   So, the sequence $(c_n)_{n\in \mathbb{N}}$ converges to $-2$.
   
4. $\displaystyle{d_n=\frac{n^2+1}{n+1}, \ n\in \mathbb{N}}$
   
   We have that\begin{equation*}\lim_{n\rightarrow \infty}d_n=\lim_{n\rightarrow \infty}\frac{n^2+1}{n+1}=\lim_{n\rightarrow \infty}\frac{n\left (n+\frac{1}{n}\right )}{n\left (1+\frac{1}{n}\right )}=\lim_{n\rightarrow \infty}\frac{n+\frac{1}{n}}{1+\frac{1}{n}}=\frac{\infty+0}{1+0}=\infty\notin \mathbb{R}\end{equation*}
   
   So, the sequence $(d_n)_{n\in \mathbb{N}}$ is not convergent. How can we check if it has convergent subsequence? (Wondering)

Is everything else correct? (Wondering)

 

[I like Serena]

Hey mathmari! (Smile)

It all looks correct to me.

We have that\begin{equation*}\lim_{n\rightarrow \infty}d_n=\lim_{n\rightarrow \infty}\frac{n^2+1}{n+1}=\lim_{n\rightarrow \infty}\frac{n\left (n+\frac{1}{n}\right )}{n\left (1+\frac{1}{n}\right )}=\lim_{n\rightarrow \infty}\frac{n+\frac{1}{n}}{1+\frac{1}{n}}=\frac{\infty+0}{1+0}=\infty\notin \mathbb{R}\end{equation*}

So, the sequence $(d_n)_{n\in \mathbb{N}}$ is not convergent. How can we check if it has convergent subsequence? (Wondering) )

Doesn't our limit calculation imply that however big we want a value of the sequence to be, we can find an $N$ such that for all $n>N$ each element $d_n$ of the sequence is bigger than that value? (Wondering)

In other words, there can be no convergent subsequence.

 

[mathmari]

It all looks correct to me.
Doesn't our limit imply that however big we want a value of the sequence to be, we can find an $N$ such that for all $n>N$ each element $d_n$ of the sequence is bigger than that value? (Wondering)

In other words, there can be no convergent subsequence.

Ah ok! Thank you very much! (Yes)