series and lim sup proof

Question

if $∑|a_n|$ is convergent and for all n $ |\frac{b_{n+1}}{b_n}|\le|\frac{a_{n+1}}{a_n}|$. (1)
prove $∑|b_n|$ is convergent.

My answer is :
since $∑|a_n|$ is convergent, then $∑a_n$ is convergent. so it must have met the ratio test, so
$\lim \sup |\frac{a_{n+1}}{a_n}|<1$
from (1) we conclude $ |\frac{b_{n+1}}{b_n}|<1$ as well. which means $∑b_n$ also meets the ratio test. hence it is absolutely convergent, so $∑|b_n|$ is convergent too.
I am confused with lim sup thing. can I say
since $ |\frac{b_{n+1}}{b_n}|\le|\frac{a_{n+1}}{a_n}|$ then $\lim \sup |\frac{b_{n+1}}{b_n}| \le \lim \sup |\frac{a_{n+1}}{a_n}|$ ( if yes, what is the proof for that? I know such proof for just limit ) Thanks

Answer 1

As pointed out in the comments your proof is not correct.

Let $c_n=\frac {|a_{n+1}|} {|a_n|}$. Then $|b_{n+1}|\leq |c_n |b_n|$ from this you get $|b_n| \leq c_1c_2...c_{n-1} |b_1|=\frac {|a_n| } {|a_1|} |b_1|$ and hence $\sum |b_n| <\infty$.

Answer 2

For convenience of notation, we can assume, without loss of generality, that $a_n,b_n > 0$, for all $n$.

With that assumption, suppose $$\frac{b_{n+1}}{b_n} \le \frac{a_{n+1}}{a_n}$$ for all $n$, and suppose $\sum a_n$ converges.

Claim:$\;\sum b_n$ converges.

Proof:$\;$Rearranging the given inequality, we get $$\frac{b_{n+1}}{a_{n+1}} \le \frac{b_n}{a_n}$$ for all $n$.

It follows that $$\frac{b_n}{a_n} \le \frac{b_1}{a_1}$$ for all $n$.

Thus, letting $c={\large{\frac{b_1}{a_1}}}$, we have $b_n\le ca_n$, for all $n$.

Hence, since $\sum ca_n$ converges, so does $\sum b_n$, as was to be shown.