# Limit comparison test proof

We had the following theorem in class:

Let $(a_n)$ and $(b_n)$ be sequences and $b_n>0$ and $\lim_{n\rightarrow\infty}\frac{a_n}{b_n}=L$ with $L\in\mathbb R\backslash\{0\}$. Then $\sum a_n$ converges if and only if $\sum b_n$ converges.

So by recapitulating the lecture I've tried to prove it but I didn't get it. It's obvious that you have to use the comparison test, but how? Can anybody help? Thanks a lot!

Hint: Suppose that $L\gt 0$ (a similar argument will deal with $L\lt 0$).
Because $\lim_{n\to\infty}\frac{a_n}{b_n}=L$, there is an $N$ such that if $n\gt N$ then $$\frac{L}{2}\lt \frac{a_n}{b_n}\lt \frac{3L}{2}.$$ The above inequality follows from the definition of limit by taking $\epsilon=\frac{L}{2}$.
• thanks. this really helped me. just one question: $\sum a_n$ converges, so $\lim a_n=0$. can you follow $b_n=0$ (then I might have one more (different) proof) – sheldoor Jan 3 '13 at 22:46
• It is true that $b_n\to 0$. But from that we cannot conclude the convergence of $\sum b_n$. And if $\lim\frac{a_n}{b_n}=0$, then from convergence or divergence of $\sum a_n$ we cannot say anything about $\sum b_n$. – André Nicolas Jan 4 '13 at 6:55