# Manipulating divergent sequences using the Algebraic Limit Theorems

$$\def\N{\mathbf{N}}$$ $$\def\R{\mathbf{R}}$$ 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?

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 $$c$$.