# real analysis – If \$ frac{b_n}{a_n} overset{ntoinfty}{longrightarrow} alphaneq0, \$ then \$sum a_n \$ converges \$ iff sum b_n \$ converges?

I’m stuck on an easy question because… I don’t know. I’m old? Anyway, it’s annoying me, because I usually don’t find these questions too difficult, and I’m pretty sure I’ve seen it before.

If $$frac{b_n}{a_n} overset{ntoinfty}{longrightarrow} alphaneq0,$$ then $$displaystylesum a_n$$ converges $$iff displaystylesum b_n$$ converges

No way this can be false, since at some point $$b_n$$ will be tied relatively closely to $$alpha a_n,$$ ( i.e. $$b_n$$ will be relatively far away from $$0$$ compared with the distance from $$b_n$$ to $$alpha a_n. )$$ I would like to solve it all in one go, i.e. avoiding splitting into cases like $$alpha < 0$$ and then $$alpha > 0,$$ because due to “$$b_n$$ will be bounded relatively closely to $$alpha a_n$$“, this seems to me like it should be straightforward.

I’ve tried a few avenues to go down with regards to a proof: mainly starting from $$varepsilon-n$$ definitions of convergence, but before I spend more time on this, I just want to check if I’m missing something really obvious.