Are my ideas as for the convergence right?

[evinda]

Hello! (Wave)

I am looking at the following exercise:

1. Let $(a_n)$ be a sequence of real numbers such that $a_{2n} \to a$ and $a_{2n+1} \to a$ for some real number $a$. Show that $a_n \to a$.
2. We are given the sequence $x_n=0$ if $n$ is even, $x_n=1$ if $n$ is odd. Check as for the convergence the sequence $\left( \frac{x_1+x_2+\dots+x_n}{n}\right)$.

I have thought the following:

1. Since $a_{2n} \to a$, we have that $\forall \epsilon>0 \ \exists n_1$ such that $|a_{2n}-a|<\epsilon, \forall n \geq n_1$.
   
   Since $a_{2n+1} \to a$, we have that $\forall \epsilon>0 \ \exists n_2$ such that $|a_{2n+1}-a|<\epsilon, \forall n \geq n_2$.
   
   Then we have that $|a_n-a|<\epsilon, \forall n \geq \max \{ n_1,n_2\}$ and $\forall \epsilon$ and thus $a_n \to a$.
   
   Is this right? (Thinking)
   
2. We have that $0 \leq \frac{x_1+x_2+\dots+x_n}{n} \leq \frac{\frac{n}{2}}{n}=\frac{1}{2}$.
   
   Does this help somehow? (Thinking)

 

[I like Serena]

Hello! (Wave)

I am looking at the following exercise:

1. Let $(a_n)$ be a sequence of real numbers such that $a_{2n} \to a$ and $a_{2n+1} \to a$ for some real number $a$. Show that $a_n \to a$.
2. We are given the sequence $x_n=0$ if $n$ is even, $x_n=1$ if $n$ is odd. Check as for the convergence the sequence $\left( \frac{x_1+x_2+\dots+x_n}{n}\right)$.

I have thought the following:

1. Since $a_{2n} \to a$, we have that $\forall \epsilon>0 \ \exists n_1$ such that $|a_{2n}-a|<\epsilon, \forall n \geq n_1$.

Since $a_{2n+1} \to a$, we have that $\forall \epsilon>0 \ \exists n_2$ such that $|a_{2n+1}-a|<\epsilon, \forall n \geq n_2$.

Then we have that $|a_n-a|<\epsilon, \forall n \geq \max \{ n_1,n_2\}$ and $\forall \epsilon$ and thus $a_n \to a$.

Is this right?

Hey evinda!

Yep. (Nod)

2. We have that $0 \leq \frac{x_1+x_2+\dots+x_n}{n} \leq \frac{\frac{n}{2}}{n}=\frac{1}{2}$.

Does this help somehow?

Don't we have:
$$\frac n2 \le x_1+...+x_n \le \frac{n+1}2$$
? (Wondering)

 

[evinda]

Don't we have:
$$\frac n2 \le x_1+...+x_n \le \frac{n+1}2$$
? (Wondering)

How do we get this inequality?

The sum $x_1+\dots+x_n$ has $n$ summands. If $n$ is even, then the half, namely $\frac{n}{2}$ summands will be equal to $1$.
If $n$ is odd, then $\frac{n-1}{2}$ summands will be equal to $1$.

Or am I wrong? (Thinking)

 

[I like Serena]

How do we get this inequality?

The sum $x_1+\dots+x_n$ has $n$ summands. If $n$ is even, then the half, namely $\frac{n}{2}$ summands will be equal to $1$.
If $n$ is odd, then $\frac{n-1}{2}$ summands will be equal to $1$.

Or am I wrong?

Consider $n=1$, which is odd, so we have $x_1=1$.
But $\frac{n-1}{2} = \frac{1-1}{2} = 0 \ne 1$, isn't it? (Thinking)

 

[evinda]

Consider $n=1$, which is odd, so we have $x_1=1$.
But $\frac{n-1}{2} = \frac{1-1}{2} = 0 \ne 1$, isn't it? (Thinking)

Oh yes, right... If $n$ is odd, then $\frac{n-1}{2}+1=\frac{n+1}{2}$ summands will be equal to $1$, right? (Thinking)

That's why we get $\frac{n}{2}\leq x_1+\dots+x_n \leq \frac{n+1}{2}$.

Then we have that $\frac{1}{2} \leq \frac{x_1+x_2+\dots+x_n}{n} \leq \frac{1}{2} \left( 1+\frac{1}{n}\right) \Rightarrow \frac{1}{2} \leq \lim_{n \to +\infty} \frac{x_1+ x_2+\dots+ x_n}{n} \leq \lim_{n \to +\infty} \frac{1}{2} \left( 1+\frac{1}{n} \right)=\frac{1}{2}$ and so the sequence $\frac{x_1+x_2+\dots+x_n}{n}$ converges to $\frac{1}{2}$.

Right?

 

[I like Serena]

(Nod)

 

[evinda]

(Nod)

Nice, thank you... (Smile)