{a_n} and {a_n+b_n} converge. Prove {b_n} converges.

**dwsmith** · June 24th 2011, 11:55 AM

$\{a_n\}$ and $\{a_n+b_n\}$ converge. Prove $\{b_n\}$ converges.

Let $\{a_n\}$ and $\{a_n+b_n\}$ converge to $A, \ A+\alpha$ , respectively,

Then there is an $\epsilon>0$ and $\exists N\in\mathbb{N}, \ n\geq N$ ,

$|a_n-A|<\frac{\epsilon}{2}$

and

$|a_n+b_n-(A+\alpha)|\leq \epsilon$

$|a_n+b_n-(A+\alpha)|\leq |a_n-A|+|b_n-\alpha|<\frac{\epsilon}{2}+|b_n-\alpha|<\epsilon$

$\Rightarrow |b_n-\alpha|<\frac{\epsilon}{2}<\epsilon$

Therefore, $\{b_n\}$ converges.

Correct?

**Joanna** · June 24th 2011, 12:06 PM

It looks good except for this part:

Then there is an $\epsilon>0$

You need to say something like "Let $\epsilon>0$ be given."
Otherwise you are saying that there exists an epsilon and N for which the following is true, which is not what you want to prove. By saying that epsilon is given and arbitrary, you essentially show that it is true for any epsilon (greater than zero, of course).

**Ackbeet** · June 24th 2011, 12:09 PM

As Joanna said, it needs to be a "for all epsilon", not a "there exists an epsilon". You need to work with

$|b_{n}-\alpha|=|b_{n}+a_{n}-a_{n}+A-A-\alpha|.$

See where that leads. Your proof is incorrect, because you're inserting items in the inequality chain: that's not valid.

**dwsmith** · June 24th 2011, 12:20 PM

Originally Posted by Ackbeet

As Joanna said, it needs to be a "for all epsilon", not a "there exists an epsilon". You need to work with

$|b_{n}-\alpha|=|b_{n}+a_{n}-a_{n}+A-A-\alpha|.$

See where that leads. Your proof is incorrect, because you're inserting items in the inequality chain: that's not valid.

What is "not valid"?

We know $\{a_n+b_n\}$ converges. So if we let it converge to $A+\alpha$ , we have $|a_n+b_n-(A+\alpha)|<\epsilon$

And by the triangle inequality, $|a_n-A|+|b_n-\alpha|\geq |a_n+b_n-(A+\alpha)|$

Since we also know $\{a_n\}$ converges, I can let it converge to some A.

Because it converges $|a_n-A|<\frac{\epsilon}{2}<\epsilon$

Since $|a_n-A|<\frac{\epsilon}{2}$ , we can substitute $\frac{\epsilon}{2}$ for $|a_n-A|$ and make the inequality strictly less than.

That yields:

$|a_n-A|+|b_n-\alpha|<\frac{\epsilon}{2}+|b_n-\alpha|$

But we also know that $|a_n+n_n-(A+\alpha)|<\epsilon$ .

So $\frac{\epsilon}{2}+|b_n-\alpha|<\epsilon$

I don't see what is wrong in the inequality structure.

**Joanna** · June 24th 2011, 12:23 PM

Originally Posted by Ackbeet

$|b_{n}-\alpha|=|b_{n}+a_{n}-a_{n}+A-A-\alpha|.$

See where that leads. Your proof is incorrect, because you're inserting items in the inequality chain: that's not valid.

Right! Missed that for some reason.

Originally Posted by dwsmith

$\{a_n\}$ and $\{a_n+b_n\}$ converge. Prove $\{b_n\}$ converges.

$|a_n+b_n-(A+\alpha)|\leq \epsilon$

$|a_n+b_n-(A+\alpha)|\leq |a_n-A|+|b_n-\alpha|<\frac{\epsilon}{2}+|b_n-\alpha|<\epsilon$

You cannot say that $\frac{\epsilon}{2}+|b_n-\alpha|<\epsilon$ . You know that $|a_n+b_n-(A+\alpha)|<\epsilon$ , but you just showed that $|a_n+b_n-(A+\alpha)|<\frac{\epsilon}{2}+|b_n-\alpha|$ so it is possible that $\frac{\epsilon}{2}+|b_n-\alpha|\geq\epsilon$ .

**dwsmith** · June 24th 2011, 12:41 PM

I don't agree with you two and here is why:

So if we let $\epsilon>0$ . Then $\exists N\in\mathbb{N}, \ n\geq N$

Now, let $\{a_n\}\to A \ \text{and} \ \{a_n+b_n\}\to A+\alpha$

Since those two sequences converge,

$|a_n-A|<\epsilon \ \text{but it is also true then that} \ |a_n-A|<\frac{\epsilon}{2}<\epsilon$

And

$|a_n+b_n-(A+\alpha)|<\epsilon$

By the triangle inequality,

$|a_n+b_n-(A+\alpha)|\leq \underbrace{|a_n-A|}_{|a_n-A|<\frac{\epsilon}{2}}+|b_n-\alpha|<\underbrace{\frac{\epsilon}{2}}_{\text{thi s is + and} \ < |a_n-A|}+|b_n-\alpha|<\epsilon$

Now, we have

$|a_n+b_n-(A+\alpha)|<\frac{\epsilon}{2}+|b_n-\alpha|<\epsilon$

Let's subtract epsilon over 2.

$|b_n-\alpha|<\frac{\epsilon}{2}$

I started with something giving to be true {a_n+b_n} converges, used another true piece {a_n} converges, used the triangle inequality, and made substitution that is true.

If $|a_n+b_n-(A-\alpha)|\leq|a_n-A|+|b_n-\alpha|$ , it is $<\frac{\epsilon}{2}+|b_n-\alpha|$ which is by definition less than epsilon.

**Ackbeet** · June 24th 2011, 12:56 PM

$\underbrace{|a_n+b_n-(A+\alpha)|\leq |a_n-A|+|b_n-\alpha|<\frac{\epsilon}{2}+|b_n-\alpha|}_{\text{I agree with all of this,}}<\epsilon.$

$|a_n+b_n-(A+\alpha)|\leq |a_n-A|+|b_n-\alpha|<\underbrace{\frac{\epsilon}{2}+|b_n-\alpha|<\epsilon}_{\text{but not this.}}.$

You're essentially saying that because A < B and A < C, that therefore A < B < C. But 4 < 5, and 4 < 6; that does not imply 4 < 6 < 5.

**Plato** · June 24th 2011, 01:07 PM

Originally Posted by dwsmith

$\{a_n\}$ and $\{a_n+b_n\}$ converge. Prove $\{b_n\}$ converges.

I know this has been answered. But here is a flowing proof.
First suppose that $(a_n)\to\alpha~\&~(a_n+b_n)\to \alpha+\beta$ .
Suppose that $\varepsilon > 0$ . Now use the definition twice.
$\left( {\exists N_1 } \right)\left[ {n \geqslant N_1 \, \Rightarrow \,\left| {a_n - \alpha } \right|} \right] < \frac{\varepsilon }{2}$
$\left( {\exists N_2 } \right)\left[ {n \geqslant N_2 \, \Rightarrow \,\left| {a_n + \beta _n - (\alpha + \beta) } \right|} \right] < \frac{\varepsilon } {2}$ .

Note that we need two values of $N$ that we need to insure both condition hold.
So let $N=N_1+N_2$ . If $n\ge N$ then
$\begin{array}{rcl}{\left| {b_n - \beta } \right|} & = & {\left| {\left( {a_n + \beta _n } \right) - \left( {\alpha + \beta } \right) + \alpha -a_n } \right|}\\{} & \leqslant & {\left| {\left( {a_n + \beta _n } \right) - \left( {\alpha + \beta } \right)} \right| + \left| {\alpha - a_n } \right|}\\{} & \leqslant & {\frac{\varepsilon }{2} + \frac{\varepsilon }{2} = \varepsilon } \\ \end{array}$

**Joanna** · June 24th 2011, 02:02 PM

Here's a specific example where your argument fails.

As Plato pointed out, you need an N for each sequence; let $N_{1}$ be such that $n>N_{1}$ implies $|a_{n}-A|<\frac{\epsilon}{2}$ and let $N_{2}$ be such that $n>N_{2}$ implies $|a_{n}+b_{n}-(A+\alpha)|<\epsilon$ . Then we can obtain both inequalities with $N=\max\{N_{1}, N_{2}\}$ .
Then n>N implies $|a_{n}-A|<\frac{\epsilon}{2}$ and $|a_{n}+b_{n}-(A+\alpha)|<\epsilon$ . This is a particular N.

Suppose that we already know that $b_{n}$ converges to $\alpha$ . Also suppose that $\alpha - b_{N+1}=\epsilon$ and $a_{N+1}-A=\frac{\epsilon}{3}$ (which is certainly less than $\frac{\epsilon}{2}$ ).

Then we have
$|a_{N+1}+b_{N+1}-(A+\alpha)|=|a_{N+1}-A-(\alpha - b_{N+1})|=|\frac{\epsilon}{3}-\epsilon|=|-\frac{2 \epsilon}{3}|<\epsilon$ but
$|a_{N+1}-A|+|b_{N+1}-\alpha|=\frac{\epsilon}{3}+\epsilon=\frac{4 \epsilon }{3}>\epsilon$ .

The point is that there exists an $n_{1}>N$ such that $|b_{n_{1}}-\alpha|=\epsilon$ and $|a_{n_{1}}+b_{n_{1}}-(A+\alpha)|<\epsilon$ . You would need to find a larger N to get $|b_{n}-\alpha|<\epsilon$ .

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