0
$\begingroup$

Prove:

Let $a_n, b_n$ be positive sequences so that $\limsup \frac {a_n}{b_n} \lt\infty$. If $\sum b_n$ converges then $\sum a_n$ converges.

So far I was thinking that from $\limsup \frac {a_n}{b_n} \lt\infty$, we know that $a_n$ doesn't have a subsequential limit of positive infinity and $b_n$ doesn't have a subsequential limit of $0$, since those two cases would make $\limsup \frac {a_n}{b_n}$ be infinity... but I'm not sure if that's what $\limsup \frac {a_n}{b_n}$ means.

If it means supremum of the set of subsequential limits of the sequences of ratios of each individual term. that means that eventually $b_n$ will be larger than $a_n$. Since otherwise it would tend to infinity.

Then we can use theorem $3.25$ from Rudin, that if $|a_n| \le c_n$ for $n \ge N$, where $N$ is some fixed integer, and $\sum c_n$ converges, then $\sum a_n$ converges as well.

Tell me what you guys think, many thanks.

$\endgroup$
  • 3
    $\begingroup$ Hint: If $L= \limsup a_n/b_n < C$, then there exists $n$ such that $\sup_{k \geqslant n}a_k/b_k < C$ $\endgroup$ – RRL Mar 28 '16 at 14:58
  • $\begingroup$ what does that notation means, sup k>n ? $\endgroup$ – mac5 Mar 28 '16 at 15:01
  • $\begingroup$ It means least upper bound of set $\{a_k: k \geqslant n\}$. Its in the definition $\limsup x_n = \inf_{n}\sup_{k \geqslant n}x_k = \lim_{n \to \infty}\sup_{k \geqslant n}x_k$. An essential property of $\limsup a_n/b_n$ is the terms of sequence are eventually less than any number greater than $\limsup a_n/b_n+ \epsilon$. This helps you compare $a_n$ to $b_n$ for sufficiently large $n$. $\endgroup$ – RRL Mar 28 '16 at 15:06

Your Answer

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

Browse other questions tagged or ask your own question.