0
$\begingroup$

Consider two sequences $\{a_n\}_{n\in \mathbb{N}}\in [0,1]$ and $\{b_n\}_{n\in \mathbb{N}}>0$ and suppose $$ \begin{cases} \lim_{n\rightarrow \infty} (a_n-b_n)=0\\ \lim_{n\rightarrow \infty} b_n=0\\ \end{cases} $$ Does this imply $\lim_{n\rightarrow \infty}a_n=0$?

$\endgroup$
3
$\begingroup$

Yes, because$$\lim_{n\to\infty}a_n=\lim_{n\to\infty}\bigl((a_n-b_n)+b_n\bigr)=\lim_{n\to\infty}(a_n-b_n)+\lim_{n\to\infty}b_n=0+0=0.$$

$\endgroup$
1
$\begingroup$

An alternative: assume towards a contradiction that $\lim_{n\rightarrow\infty} a_n = c$ for some $c\neq 0$

Then we'd have $\lim_{n\rightarrow\infty} (a_n - b_n) = \lim_{n\rightarrow\infty} a_n - \lim_{n\rightarrow\infty}b_n = c - 0 = c\neq 0$

Which is a contradiction. All I've skipped over is the prove that "a limit of a sum of sequences is the sum of the limit of the sequences". We're assuming both exist and are finite.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.