We need to proof that $b_n/a_n\rightarrow0$ for $a_n\rightarrow\pm\infty$ and $(b_n)$ is restricted. But I came to $|b_n| \cdot 1/|a_n| \lt \epsilon$ and now I'm really stuck.. Can someone help me?
Limit of quotient is zero if denominator goes to infinity and numerator is bounded
4
$\begingroup$
real-analysis
sequences-and-series
analysis
limits
-
2hint: Let us suppose for the moment that $a_n\to+\infty$ Use the squeezing theorem. Since $(b_n)$ is bounded, there exists a costant M>0 such that -M
, for every $n$ but then -M/a_n ... For the case $a_n\to-\infty$ there is a change in the sign of the inequalities, but not much more than that. Can you go on from here? – 2011-09-23
1 Answers
4
Let $\varepsilon > 0$. Since $(b_n)$ is bounded there exists a constant $M$ such that $|b_n| < M$ for all $n$. Moreover $|a_n| \to \infty$ hence for some $N$ we get $|a_n| > M/\varepsilon$ for $n > N$. It is equivalent to $|1/a_n| < \varepsilon/M$.
Finally we obtain $\Bigg|\frac{b_n}{a_n}\Bigg| < \varepsilon$ for $n > N$ which means that $b_n/a_n \to 0$.
-
0@Gortaur: Oh, of course. Thanks. – 2011-09-23