Exercise 2.3.8

An excellent exercise to develop your intuition on convergence vs divergence is exercise 2.3.8. It's also excellent exam material! It has four parts - perhaps, someone could type them up here?

Exercise 2.3.8 states:
Give an example of each of the following, or state that such a request is impossible by referencing the proper theorem(s):

Here are the (five) parts of this problem:
a) sequences $(x_n)$ and $(y_n)$, which both diverge, but whose sum $(x_n + y_n)$ converges
b) sequences $(x_n)$ and $(y_n)$, where $(x_n)$ converges, $(y_n)$ diverges and $(x_n +y_n)$ converges
c) a convergent sequence $(b_n)$ with $(b_n) \neq 0$ for all such $n$ that $(\frac{1}{b_n})$ diverges
d) an unbounded sequence $(a_n)$ and a convergent sequence $(b_n)$ with $(a_n - b_n)$ bounded
e) two sequences $(a_n)$ and $(b_n)$, where $(a_nb_n)$ and $(a_n)$ converges but $(b_n)$ does not

Just gonna do one so other people have a chance to respond as well but...

a.) If we have $(x_n) = (-1)^n$ and $(y_n)= -(-1)^n$.
The output of each sequence is $(x_n) = (-1, 1, -1, 1, ... )$ and $(y_n) = (1, -1, 1, -1, ... )$ so you can see they both clearly diverge because they only ever alternate between the same two values. But added together, they converge to zero because each term in $(x_n)$ cancels out each term in $(y_n)$.

c) For this one consider $b_n = (1, \frac{1}{2}, \frac{1}{3}, \dots ,\frac{1}{n}, \dots)$. Note that $b_n\rightarrow 0$ and $b_n\neq 0$ for all $n$. However, note $\frac{1}{b_n} = (1, 2, 3, \dots, n, \dots)$ diverges.

e) For this one consider the sequences $a_n = (1,\frac{1}{2},\frac{1}{3},\dots,\frac{1}{n},\dots)$ and $b_n = (-1,1,-1,\dots,(-1)^n,\dots)$. Note $a_n\rightarrow 0$ and $b_n$ diverges. We also find $a_nb_n=(-1,\frac{1}{2},-\frac{1}{3},\dots,(-1)^n\frac{1}{n},\dots)$ converges to 0. Therefore, the correct properties are satisfied.

I'm going to try the two that aren't posted:
(b) By the Algebraic Limit Theorem this can't be a thing…
(d) This one is also impossible, because every convergent sequence is bounded.

Regarding (b), can we use the algebraic limit theorem if one sequence is not convergent?

In the definition of the Algebraic Limit Theorem we need a $lima_{n}=a$ AND a $limb_{n}=b$. Otherwise we can't assert that $lim(a_{n}+b_{n})=a+b$. Unless I misunderstood your question...

That's what I suspected. In that case, I am not entirely sure we can conclude (b) is false using only the algebraic limit theorem.

Can you give an example of when it would work? Or do you have another suggestion?

After digging around a bit, I see how it does give us that (b) is impossible.

Say that $(x_n) \rightarrow x$ and $(x_n+y_n) \rightarrow L$, then $(x_n+y_n - x_n) = y_n$ must converge to $L-x$.

I hadn't thought of that trick and got stuck thinking only about one part of the algebraic limit theorem. Anyway, thanks for your patience with me!

We need Discourse for problems in mathematics too.

When I was working on this I figured I would try something like you did, qkhan. But we could totally generalize this answer a bit.

So for a)
If we have $(x_n) = (-a)^n$ and $(y_n) = -(-a)^n$, with $a \in \mathbb N$ then the outputs will be divergent, but every $(x_n + y_n)$ would converge because of the same thing.

Oh that's clever, I didn't even think of that! I like that. $(-1)^n$ is the most simplistic and easy to visualize sequence for most people, but you're totally right, you can easily extend that out. Nice, thanks for sharing.