$ \lim_{n\rightarrow \infty}[a_n-a]=0 $ implies $ \lim_{n\rightarrow \infty}\frac{1}{n}[a_n-a]=0 $?

Question

Suppose that $$ \lim_{n\rightarrow \infty}[a_n-a]=0 $$ Does this imply $$ \lim_{n\rightarrow \infty}\frac{1}{n}[a_n-a]=0 $$ ? Does the answer change if in place of $a$ we have another sequence $b_n$?

Answer 1

Yes. We have that $\lim_{n\to\infty}[a_nb_n]=\lim_{n\to\infty}a_n\lim_{n\to\infty}b_n$ if the limits on the right hand side exist. Hence, $$ \lim_{n\to\infty}\frac1n[a_n-a]=\lim_{n\to\infty}\frac1n\lim_{n\to\infty}[a_n-a]=0\cdot0=0. $$ If we replace $a$ with some sequence, the limit might not be $0$ anymore (take, for example, $a_n=1/n$ and instead of $a=0$ take $b_n=n$).

Answer 2

For constant $a$:

$$ \lim_{n\rightarrow \infty}\frac{1}{n}[a_n-a] $$

$$ =\lim_{n\rightarrow \infty}\frac{1}{n}\lim_{n\rightarrow \infty}[a_n-a]=(0)(0) $$

For sequence $b_n$:

$$ \lim_{n\rightarrow \infty}\frac{1}{n}[a_n-b_n] $$

$$ =\lim_{n\rightarrow \infty}\frac{1}{n}\lim_{n\rightarrow \infty}[a_n-b_n]=(0)(0) $$

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Remark: Wherever they exist, limits of sequences behave like limits of functions, i.e. We can split of limits by products into products of limits. This is because a sequence is a function whose domain is countable.