If $\frac{a_n}{b_n} \rightarrow 1$ does $a_n$ converge?

Question

$\frac{a_n}{b_n} \rightarrow 1$ and $\sum_{n=1}^\infty b_n$ converges, can it be concluded that $\sum_{n=1}^\infty a_n$ converges?
My attempt at an answer to this question: since $\sum_{n=1}^\infty b_n$ converges, $b_n \rightarrow 0$. Because of this, $a_n \rightarrow 0$ equally fast. However, I'm well aware that this does not imply that $\sum_{n=1}^\infty a_n$ converges. I'm stuck at that point, though, as I'm not sure what other conclusions can be drawn. Could anyone help me out?

Answer 1

I actually think the answer is no. Take $b_{n} = \frac{(-1)^{n}}{\sqrt{n}}$ and $a_{n} = \frac{(-1)^{n}}{\sqrt{n}} + \frac{1}{n}$. Then $\sum_{n}b_{n}$ converges and $\sum_{n}a_{n}$ diverges but $\frac{a_{n}}{b_{n}} \rightarrow 1$ as

Answer 2

This is the limit comparison test. As long as $\sum b_n$ converges, the limit of $\frac{a_n}{b_n}$ being any real number is enough to guarantee that $\sum a_n$ converges.

Indeed, since $\sum b_n$ is convergent, then so is $\sum kb_n$ for any real $k$. Whatever limit you obtain for $\frac{a_n}{b_n}$, choose some $k$ larger than that, and then look at a direct comparison between $\sum a_n$ and $\sum kb_n$.

(I'm assuming in this discussion that all terms are positive. If not, then you may have to make some adjustments.)

Answer 3

$$a_n = \frac{a_n}{b_n}\cdot b_n$$

(given that $b_n\neq 0$) The intuition now tells us, that from a certain $N$ on, $\frac{a_n}{b_n}$ will be so close to $1$ that $a_n$ basically add not much more than $b_n$, so if $\sum b_n$ converges, so will $\sum a_n$.

But as the other answers have raised concern if intuition is not enough in this case, so let's see what exactly we need. We can safely assume that $b_n\neq 0$ eventually, e.g. starting from $N_1\in\mathbb{N}$ (else the limit of $\frac{a_n}{b_n}$ couldn't be defined). Set $d=\lim_{n\to\infty}\frac{a_n}{b_n}$ ($d=1$ in our case). Let $c$ be any positive real (especially $0<c<|d|$), then we have $d-c < \frac{a_n}{b_n} < d+c$ eventually, e.g. starting from $N_2$. Then we have $$(d-c)b_n < a_n < (d+c)b_n$$ So with $N=\max\{N_1,N_2\}$ in mind, we can say that $(d-c)b_n < a_n < (d+c)b_n$ holds eventually Here are some facts:

• given $t\in\mathbb{R}$ with $t\neq 0$: $\sum b_n$ converges iff $\sum tb_n$ converges
• given that $|a_n|<(d+c)b_n$ eventually, $\sum(d+c)b_n$ converges implies $\sum a_n$ converges
• given on the other hand $a_n > (d-c)b_n \geq 0$ eventually, $\sum(d-c)b_n$ diverges implies $\sum a_n$ diverges

So if $d$ is positive (or even $1$ in our case) and if $b_n$ is eventually positive, then we can say that $\sum a_n$ converges if and only if $\sum b_n$ converges. If $b_n$ is not eventually positive, counterexamples can be constructed.