Manipulating divergent sequences using the Algebraic Limit Theorems

Question

$\def\N{\mathbf{N}}$ $\def\R{\mathbf{R}}$ Note: We're working in the metric space $\R$.

Back to basics on this one. I'm trying to prove the following proposition using what I know about the algebraic limit theorems (ALTs) for sequences:

Alegbraic Limit Theorems (stated in brief). The limit of the sum/product/quotient of any two convergent sequences is the sum/product/quotients of the limit of each sequences, provided the denominator sequence in the quotient case does not converge to $0$ or any of those terms $=0$.

Proposition. Let $(a_n)\to c$ and $(b_n)\to c$ be two sequences such that both converge to some common limit $c$. Now let $(q_n), (\lambda_n)$ be two more sequences such that for each $n\in\N$, $$q_n= b_n\lambda_n+ a_n(1-\lambda_n)$$ Then $(q_n)$ also converges to $c$.

Proof. We must have \begin{align}\label{test} \lim_{n\to\infty}{q_n} &= \lim_{n\to\infty}{\big(b_n\lambda_n + a_n(1-\lambda_n)\big)}\tag{1} \\ &= \lim_{n\to\infty}{b_n\lambda_n} + \lim_{n\to\infty}{a_n(1-\lambda_n)} \tag{2}\\ &= \lim_{n\to\infty}{b_n}\cdot \lim_{n\to\infty}{\lambda_n} + \lim_{n\to\infty}{a_n}\cdot \lim_{n\to\infty}{(1-\lambda_n)}\tag{3}\\ &= c\cdot \lim_{n\to\infty}{\lambda_n} + c\cdot \lim_{n\to\infty}{(1-\lambda_n)}\tag{4}\\ &= c\lim_{n\to\infty}{\big(\lambda_n+(1-\lambda_n)\big)}\tag{5} \\ &= c\cdot \lim_{n\to\infty}{1}\tag{6}\\ &=c\tag{7}\\ \end{align} Therefore, $(q_n)$ converges to $c$. $\square$

High-school-me would naively look at these strings of equations and be satisfied with this proof. Learning Real Analysis makes me come to worry (paranoically) whether the manipulations I have done are indeed justified. The issue I'm facing is that at the core of my argument, I think I'm implicitly assuming that $(\lambda_n)$ is convergent, which would then justify equations $(2), (3)$ and$(5)$, by the ALTs. But $(\lambda_n)$ could also be divergent, so most of these equations wouldn't make any sense then in that case, no? The ALTs from what I understand can be applied only to two convergent sequences, and are silent in the case of divergent sequences.

So what I think I've managed to prove, is that with the additional hypothesis that $(\lambda_n)$ is convergent, then $q_n$ converges to $c$. But intuitively I think that the additional hypothesis is not required. What, then is the issue in the argument/my own understanding of the argument, and is this a reasonable worry to have in the name of ensuring that all results must be proved carefully and rigorously?

Answer 1

You are indeed assuming $(\lambda_n)_n$ is convergent. It is ‘not okay’ to factor out $\lim_{n\to\infty}\lambda_n$ since you don’t know that this exists. Note that algebra with $\infty$ is meaningless, so there is a genuine room for error.

Let me rewrite: $$|q_n-a_n|=|b_n-a_n|\cdot|\lambda_n|$$

The sequence $(q_n)$ converges to $c$ iff. both sides of the above equation tend to zero. Although we do have $|b_n-a_n|\to0$, if $\lambda_n$ increases in magnitude sufficiently quickly it is possible that $|\lambda_n|\cdot|a_n-b_n|$ does not converge to zero. Note that if $(\lambda_n)_n$ is bounded, there is no issue. We need an unbounded divergent sequence to get a counterexample.

Consider $c=1$ and $a_n:=1-1/n,b_n:=1+1/n$. If $\lambda_n:=n$, we get: $$q_n-a_n=2$$For all $n$. So $(q_n)_n$ converges but it does not converge to $c$. If I let $\lambda_n:=n^x$ for any $x>2$, the $(q_n)_n$ won’t converge: if I let $x<2$, then they will and they will converge to $c$.