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Question_

Suppose that $a_n>0$ and $b_n>0$ for $n\ge N$ where $N$ is an integer. Given that $\lim_{n \to \infty} {a_n \over b_n}=\infty$ and $\sum a_n$ converges, can anything be said about $\sum b_n$ ?

I've found the case of $\sum b_n$ converging while $\sum a_n$ converges: $$a_n={1\over n^2}, b_n={1\over n^3}$$ (Easy:D) I've also tried to find the case of $\sum b_n$ diverges, but unfortunately, I still have no idea. Nevertheless, I believe that there's no case for $\sum b_n$ diverging. The reason is as follows. When we use the definition of the limits: $$\forall M>0, \exists N \space s.t. \space \forall n\ge N \implies a_n/b_n >M$$ So, $a_n\gt Mb_n$. Due to this, $$\sum_{n=N}^\infty a_n \gt M\sum_{n=N}^\infty b_n$$ Thus, $\sum_{n=N}^\infty b_n$ converges. I thought the part $n=1 \to N$ does not affect the convergence of $\sum_{n=N}^\infty b_n$ since the number of terms is finite.

I'm asking the question to check whether the approach I've made is correct or have some contradictory points. Also, I'm wondering if there exists a sequence $b_n$ that $\sum b_n$ diverges. Thanks for giving your ideas!

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  • $\begingroup$ Your approach is fine, though it's not clear to me what it means to say "I've found the case of $\sum b_n$ converging." $\endgroup$ Commented Feb 17, 2020 at 15:46
  • $\begingroup$ Your proof is the correct idea. The only thing I would add is that you need to choose $N$ so that $a_n>0$ and $b_n>0$. Otherwise, the inequality, $a_n>M b_n$ isn't obvious. $\endgroup$
    – ProfOak
    Commented Feb 17, 2020 at 15:59

2 Answers 2

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HINT:

For large $n$ we have $a_n/b_n>1$, hence $a_n>b_n$.

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  • $\begingroup$ That's simple but is important to this question. Thanks:D $\endgroup$
    – ToBY
    Commented Feb 17, 2020 at 16:00
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$a_n,b_n>0$;

$\lim_{n \rightarrow \infty} (a_n/b_n)=\infty$:

For $M >0$, real, there is a $n_0$ s.t for $n\ge n_0$:

$ a_n/b_n >M$;

Then $a_n >Mb_n$ for $n\ge n_0$.

And now ? (Comparison test)

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  • $\begingroup$ Yeah, that's what I did. Thanks :D $\endgroup$
    – ToBY
    Commented Feb 17, 2020 at 15:59
  • $\begingroup$ ToBy. Then all is well :) $\endgroup$ Commented Feb 17, 2020 at 16:00

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