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 $\lim_{n\rightarrow \infty}a_n=0$?
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2 Answers
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Yes, because$$\lim_{n\to\infty}a_n=\lim_{n\to\infty}\bigl((a_n-b_n)+b_n\bigr)=\lim_{n\to\infty}(a_n-b_n)+\lim_{n\to\infty}b_n=0+0=0.$$
answered May 11, 2018 at 18:48
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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.
