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I have this question:

For $\large \ a_n > 0 $ and $\large \ b_n > 0 $, and $\large \lim_{n\rightarrow \infty} \frac{b_n}{a_n} = 0 $ prove that $\lim_{n\rightarrow \infty} a_n = \infty$ if $\lim_{n\rightarrow} b_n = \infty$

One way I thought I can prove this is by showing that since $\large \lim_{n\rightarrow \infty} \frac{a_n}{b_n}$ then by using the ratio test, we have:

$\large \lim_{n\rightarrow\infty} \left|\frac{a_n+1}{b_n+1}*\frac{b_n}{a_n} \right| < 1 $

and for large values of n, we have $\large \ a_{n+1} *b_n < b_{n+1}*a_n $

now since $\large \lim_{n\rightarrow \infty} b_n = \infty $ then $\large b_n$ is increasing and $\large b_n < b_{n+1} $ in all cases. This shows that $\large a_n$ can be zero, constant, or going to $\large \infty$ as well.

Logarithm
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2 Answers2

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Perhaps the easiest way to do this is as follows : Since $\lim_{n\to\infty} \frac{b_n}{a_n} = 0$, the sequence is bounded, so there exists $0\neq M \in \mathbb{R}$ such that $$ |\frac{b_n}{a_n}| \leq M $$ and hence $$ |b_n| \leq M|a_n| $$ Since $|b_n| = b_n \to \infty$, it follows (by the Squeezing principle?), that $$ \lim_{n\to\infty} M|a_n| = +\infty $$ Since $M\neq 0$, $a_n$ must also go to $\infty$.

Prahlad Vaidyanathan
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Suppose $a_n\not\rightarrow \infty$, then passing to a subsequence we can assume $a_n$ are bounded by $M$, which implies

$$\frac{b_n}{a_n} \ge \frac{b_n}{M} \rightarrow \infty$$

a contradiction.

Deven Ware
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