Suppose $\langle a_n\rangle$ and $\langle b_n\rangle$ are two convergent sequences of real numbers such that $a_n > 0$ and $b_n > 0$ for all n. Suppose $\lim_{n \to \infty} a_n = a$ and $\lim_{n \to \infty} b_n = b$. Let $c_n = \frac{a_n}{b_n}$. Then

  1. $\langle c_n\rangle$ converges if $b \gt 0$

  2. $\langle c_n\rangle$ converges only if $a = 0$

  3. $\langle c_n\rangle$ converges only if $b \gt 0$

  4. $\limsup_{x \to \infty}c_n = \infty$ if $b = 0$

My Attempt:

Since $\langle a_n\rangle$ and $\langle b_n\rangle$ are positive terms sequences so $\lim_{n \to \infty}a_n = a \gt 0$ and $\lim_{n \to \infty}b_n = b \gt 0$ Also $\langle a_n\rangle$ and $\langle b_n\rangle$ are convergent. So $a$ and $b$ are finite. Therefore $\lim_{n \to \infty}c_n = \frac{a}{b} \gt 0$ and finite. Hence using above information, we say that option 1 is true and options 2,3 are false. Also we know that if $\lim_{n \to \infty}c_n = L$ then $\limsup_{n \to \infty} = L$ and $\liminf_{n \to \infty} = L$. Hence option 4 is false. I'm right ? If not then answer my question or point out my mistake. Thanks in advance.

  • $\begingroup$ What if $b_n = \frac{1}{n}$? $\endgroup$
    – Joe
    Jun 6, 2021 at 4:24
  • 1
    $\begingroup$ A general technique that I like to use for many math questions: before trying to prove anything, try out a few examples. What if $a_n=b_n=\frac{1}{n}$? What if $a_n=b_n=1$? Ideally, I'd try to think of possible counter-examples before trying to prove one way or the other. $\endgroup$
    – Joe
    Jun 6, 2021 at 4:32
  • $\begingroup$ Writing "$\color{red}{+}\infty$" for limits over $\mathbb{N}$ doesn't make much sense. $\endgroup$
    – metamorphy
    Jun 6, 2021 at 5:46
  • $\begingroup$ "Since ⟨an⟩ and ⟨bn⟩ are positive terms sequences so limn→+∞an=a>0 and limn→+∞bn=b>0" It is absolutely not true that if every terms is not zero then a limit can not be $0$. The limit is what the sequence gets close to and need not be or have any properties of the terms. It is VERY possible for all the terms to be larger than $0$ yet have the terms get close to $0$ as the limit. Take $b_n = \frac 1n$ or $b_n = \frac 1{2^n}$. Those have limits equal to $0$ but none of the terms actually are $0$. $\endgroup$
    – fleablood
    Jun 6, 2021 at 5:47

1 Answer 1


Strict inequalities are transformed to loose inequalities by taking the limits.

Therefore $a_n,b_n>0$ implies only $a,b\ge 0$ and zero cannot be excluded.

Let start with the case $b=0$.

  • $a_n=\dfrac 1n$ and $b_n=\dfrac 1{n^2}$ then $c_n=n\to\infty$ so $(2)$ is not true.

  • $a_n=\dfrac 3n$ and $b_n=\dfrac 1n$ then $c_n=3\to 3$ so $(3)$ and $(4)$ are not true.

It remains the case $b>0$

  • In that case $c_n\to \dfrac ab$ is well defined and $(1)$ is true.

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