Let's suppose that both sequences are convergent

[DivGradCurl]

Consider the following statement:

If $$\left\{ a_n \right\}$$ and $$\left\{ b_n \right\}$$ are divergent, then $$\left\{ a_n b_n \right\}$$ is divergent.

I need to decide whether it is true or false, and explain why. The real problem is that I checked the answer in my book; it's false, but I don't understand it. Here is what I think:

Let's suppose that both sequences are convergent. Then, it follows that

$$\lim _{n\to \infty} a_n \cdot \lim _{n\to \infty} a_n = \lim _{n\to \infty} \left( a_n b_n \right) \tag{1}$$​

But, the truth is that both are divergent. So, $$\lim _{n\to \infty} a_n \neq 0$$ and $$\lim _{n\to \infty} b_n \neq 0$$. If neither is zero, then how can $$\lim _{n\to \infty} \left( a_n b_n \right) = 0$$ (so that the statement is false)? It doesn't sound reasonable if you consider (1).

Can anybody please help me clarify this?

Thank you very much.

 

[quasar987]

If a sequence is convergent, it doesn't decessarily means it converges towards 0, like you seemed to be implying in your post.

To decide wheter the proposition it's true or false, a simple counter exemple suffice. Consider $a_n = (-1)^n$ and $b_n = (-1)^n$. These are both divergent series because when n is pair, $a_n = 1$ and when n is odd $a_n = -1$, such that the limit is dependent upon n ==> it is not unique ==> it doesn't exist ==> the sequences diverge. But $a_n b_n = (-1)^{2n} = 1 \ \forall n \in \mathbb{N}$ is a sequence that converges towards 1.

Or take $a_n = (-1)^n$ and $b_n = (-1)^n+1$. Then $a_n b_n = (-1)^{2n+1} = -1 \ \forall n \in \mathbb{N}$, which converges towards -1.

 

[quasar987]

N.B. But in the case where $a_n$ and $b_n$ are divergent because they increase or decrease without limit (i.e. because their limit is plus or minus infinity), then it is true that $a_n b_n$ is also a divergent sequence.

 

[DivGradCurl]

Oh... I see what you mean. I thought that way because I had in mind the theorem that says that when a series is convergent, the terms go to 0. It doesn't apply in this case, since we only have sequences. Thanks.

 

[DivGradCurl]

I see, so it isn't true in general.

 

[NateTG]

Consider the following statement:

If $$\left\{ a_n \right\}$$ and $$\left\{ b_n \right\}$$ are divergent, then $$\left\{ a_n b_n \right\}$$ is divergent.

I need to decide whether it is true or false, and explain why. The real problem is that I checked the answer in my book; it's false, but I don't understand it. Here is what I think:

The following is a problem step:

Let's suppose that both sequences are convergent.

The hypothesis is that both sequences are divergent, not convergent, so this thought does not apply to this question in any useful way.

For example, $a_n=b_n=(-1)^n$ are two divergent sequences, but $\{a_nb_n\}$ is constant, so it clearly converges.

By the way, the theorem about series on indicates that terms in a convergent series go to zero. There are divergent series that go to zero like the harmonic series:
$$\sum_{i=1}^{\infty} \frac{1}{i}$$
And, since
$$\sum_{i=1}^{\infty} \frac{1}{i^2} = \frac{\pi^2}{6}$$
is convergent, the above is false for series as well as sequences.