# Analysis: Please check my proof on the arithmetic of CONVERGENT SEQUENCES

• October 25th 2010, 10:05 AM
SunRiseAir
Analysis: Please check my proof on the arithmetic of CONVERGENT SEQUENCES
I am new to analysis and am having a lot of trouble with this proof. I am not sure if what I have written is valid. Please let me know. Thank you.

Theorem: If {an} converges to A and {an + bn} converges to A + B, then {bn} converges.

Proof:

Since {an + bn} converges to A + B, for every ε> 0 there exists an N such that for
n > N, |(an + bn) - (A + B)| < ε

=>
|(an - A) + (bn - B)| < ε
It is true that
|(an - A) + (bn - B)| ≤ |an - A| + |bn - B|
It is also true that
|(an - A) + (bn - B)| - |an - A| < ε for n > N
Thus it is true that
|(an - A) + (bn - B)| - |an - A| ≤ |an - A| + |bn - B| - |an - A| < ε
for n > N
Therefore
|an - A| + |bn - B| - |an - A| < ε for n > N
so
|bn - B| < ε for n > N
Since for ε there exists an N such that n > N => |bn - B| < ε, then {bn} converges.■

I used the same N for {bn} that I used for {an +bn} because if it works for
{an + bn} and {bn} is inside {an + bn}, then I reason it will work for {bn} too.
• October 25th 2010, 10:48 AM
Plato
First, one can never assume that same N works in all cases.

Now you know that $|W|\le |W-U|+|U|$.
Let $U=A-a_n~\&~W=b_n-B$.
You can controle both $|A-a_n|~\&~|a_n+b_n-(A+B)|$ making each less than $\frac{\epsilon}{2}$.
• October 25th 2010, 01:52 PM
HappyJoe
Do you already know the theorems that if $\{x_n\}$ and $\{y_n\}$ converges, then $\{cx_n\}$ converges for any constant $c$, and $\{x_n+y_n\}$ converges?

In that case, you can take a short-cut. Since $\{a_n\}$ converges, you know that $\{-a_n\}$ converges. Then, since $\{a_n+b_n\}$ converges, you know that $\{a_n+b_n+(-a_n)\} = \{b_n\}$ converges.
• October 25th 2010, 02:58 PM
SunRiseAir
Quote:

Originally Posted by HappyJoe
Do you already know the theorems that if $\{x_n\}$ and $\{y_n\}$ converges, then $\{cx_n\}$ converges for any constant $c$, and $\{x_n+y_n\}$ converges?

In that case, you can take a short-cut. Since $\{a_n\}$ converges, you know that $\{-a_n\}$ converges. Then, since $\{a_n+b_n\}$ converges, you know that $\{a_n+b_n+(-a_n)\} = \{b_n\}$ converges.

Is that essentially letting $\{a_n+b_n\}$ = $\{x_n\}$, and $\{-a_n\}$ = $\{y_n\}$ in the sequence $\{x_n+y_n\}$, which is known by theorem to converge?

Quote:

Originally Posted by Plato
First, one can never assume that same N works in all cases.

Now you know that $|W|\le |W-U|+|U|$.
Let $U=A-a_n~\&~W=b_n-B$.
You can controle both $|A-a_n|~\&~|a_n+b_n-(A+B)|$ making each less than $\frac{\epsilon}{2}$.

$\{a_n\}$ converges to A and $\{b_n\}$ converges to B, prove that $\{a_n+b_n\}$ converges to A + B

you use the exact same thing:

$|a_n- A|$ < ε/2

and

$|b_n-B|$ < ε/2

So if in this proof $|b_n-B|$ < ε/2, how can $|a_n+b_n-(A+B)|$ also be less than ε/2 for the proof I'm trying to understand in the original post? Because isn't that essentially saying that all three sequences are less than ε/2?

Well, what I mean is, I see how all three could, since epsilon is arbitrary, but I don't see how it helps me. (I do realize this is because of me, not you, so don't get the wrong idea. I'm just a little lost here.)

Thanks for both of your respective input!
• October 25th 2010, 03:16 PM
Plato
If $(a_n)\to A~\&~(a_n+b_n)\to A+B$ then we have.
$\left( {\exists N_1 } \right)\left[ {n \geqslant N_1 \, \Rightarrow \,\left| {a_n - A} \right| < \varepsilon /2} \right]$ and $\left( {\exists N_2 } \right)\left[ {n \geqslant N_2 \, \Rightarrow \,\left| {\left( {a_n + b_n } \right) - \left( {A + B} \right)} \right| < \varepsilon /2} \right]$

Now let $N=N_1+N_2$ if $n\ge N$ then $n>N_1~\&~n>N_2$ so both of the above conditions are true.

Can you finish?
• October 25th 2010, 04:28 PM
SunRiseAir
Quote:

Originally Posted by Plato
If $(a_n)\to A~\&~(a_n+b_n)\to A+B$ then we have.
$\left( {\exists N_1 } \right)\left[ {n \geqslant N_1 \, \Rightarrow \,\left| {a_n - A} \right| < \varepsilon /2} \right]$ and $\left( {\exists N_2 } \right)\left[ {n \geqslant N_2 \, \Rightarrow \,\left| {\left( {a_n + b_n } \right) - \left( {A + B} \right)} \right| < \varepsilon /2} \right]$

Now let $N=N_1+N_2$ if $n\ge N$ then $n>N_1~\&~n>N_2$ so both of the above conditions are true.

Can you finish?

I don't know. I'll try. If you don't mind, criticism would be nice. I know that I suck, so hopefully I won't infuriate you with my denseness.

Since $N=N_1+N_2$, then

$|a_n- A|$ + $|a_n+b_n-(A+B)|$ $<\varepsilon /2 + \varepsilon /2 = \varepsilon$

So then

$|a_n- A|$ + $|a_n+b_n-(A+B)|$ $|a_n- A|$ + $|a_n- A|$ + $|b_n- B|$ $<\varepsilon /2 + \varepsilon /2 = \varepsilon$

So

$|a_n- A|$ + $|a_n+b_n-(A+B)|$ $2|a_n- A|$ + $|b_n- B|$ $<\varepsilon /2 + \varepsilon /2 = \varepsilon$

This should be satisfied when

$|b_n- B|$ $<\varepsilon/2$

and

$|a_n- B|$ $<\varepsilon/2$

$|a_n- B|$ $<\varepsilon/2$

then

$|b_n- B|$ $<\varepsilon/2$

for

$n\ge N$

therefore

$\{b_n\}$ converges.
• October 25th 2010, 11:38 PM
HappyJoe
Quote:

Originally Posted by SunRiseAir
Is that essentially letting $\{a_n+b_n\}$ = $\{x_n\}$, and $\{-a_n\}$ = $\{y_n\}$ in the sequence $\{x_n+y_n\}$, which is known by theorem to converge?

Yeah, that's exactly it.