Problem with convergent sequences

[gilabert1985]

Hi, I have the following problem and have done the first two questions, but I don't know how to solve the last two. Thanks for any help you can give me!

Homework Statement

Let $a_{n}\rightarrow a$, $b_{n}\rightarrow b$ be convergent sequences in $\Re$. Prove, or give a counterexample to, the following statements:

A) $a_{n}$ is a monotone sequence;
B) if $a_{n}>b_{n}+1/(n^3+4)$, then $a>b$;
C) if $a_{n}>((n^3+1)/(2n^3+1))b_{n}$, then a>b;
D) if $s_{n}=(1/n)(a_1+...+a_n)$, then $s_n \rightarrow a$.

Homework Equations

The Attempt at a Solution

I have solved the first two. For A I have given the counterexample $a_n=sin(n)/n$ and for B I have used the fact that as n goes to infinity, $1/(n^3+4)$ approaches 0, which would give $a_n > b_n$, which is a>b when n goes to infinity.

I have tried the same thing with C, but it gives me $a>(1/2)b$, which doesn't lead me anywhere, I think. And for D, I think that as n goes to infinity, $s_n$ will be close to $a_n$ because $s_n ≈ (1/n)*n*a_n$, which is the same as saying $s_n \rightarrow a$. However, I don't know if this is correct, and if it is, how am I supposed to express it?

Thanks a lot!

 

[LCKurtz]

Hi, I have the following problem and have done the first two questions, but I don't know how to solve the last two. Thanks for any help you can give me!

Homework Statement

Let $a_{n}\rightarrow a$, $b_{n}\rightarrow b$ be convergent sequences in $\Re$. Prove, or give a counterexample to, the following statements:

A) $a_{n}$ is a monotone sequence;
B) if $a_{n}>b_{n}+1/(n^3+4)$, then $a>b$;
C) if $a_{n}>((n^3+1)/(2n^3+1))b_{n}$, then a>b;
D) if $s_{n}=(1/n)(a_1+...+a_n)$, then $s_n \rightarrow a$.

Homework Equations

The Attempt at a Solution

I have solved the first two. For A I have given the counterexample $a_n=sin(n)/n$ and for B I have used the fact that as n goes to infinity, $1/(n^3+4)$ approaches 0, which would give $a_n > b_n$, which is a>b when n goes to infinity.

Think about that last statement if $a_n = \frac 2 {n^3+4}$ and $b_n=0$.

 

[gilabert1985]

Thanks!

So that would give me $2/(n^3+4)>1/(n^3+4)$, which holds when n goes to infinity.

So could I use something similar for part C then, something like $b_n=0$ or $b_n=1$? With $b_n=1$, I could have $a_n=(2n^3+1)/(2n^3+1)$?

 

[gilabert1985]

However, they would not hold for negative values of n? I am confused :/

 

[LCKurtz]

Think about that last statement if $a_n = \frac 2 {n^3+4}$ and $b_n=0$.

Thanks!

So that would give me $2/(n^3+4)>1/(n^3+4)$, which holds when n goes to infinity.

I'm not sure you understand my example and I don't know what you mean when you say it "holds when n goes to infinity". The $a_n$ and $b_n$ in my example go $a=0$ and $b=0$ respectively. The $a_n>b_n$ does not hold in the limit.