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Let $a_n$ be a nonnegative and nondecreasing sequence and $\lim a_n = \infty$, i.e. $a_n \uparrow \infty$ (1), also we know that $\lim \frac{a_{n+1}}{a_n} = 1$ (2). Find such sequence which also satisfies $\lim a_{2n}/a_n = \infty$. (3)

I tried hard, but failed to find such sequence. Then I think of trying to prove that if $\lim\frac{a_{n+1}}{a_{n}}=1$ (2), then $\lim\frac{a_{2n}}{a_n}=c$ for some constant $c\geq 1$ (4). But unfortunately I fail to prove it.

So to sum up my question: Find a nonnegative sequence satisfying (1),(2),(3). Or prove that if a nonnegative sequence satisfies (1), (2), we must have (4).

Really appreciate for any comment.

3 Answers 3

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What about $(a_n)_n$ defined by $a_1=1$ and $$ a_{n+1} = \left(1+\frac{1}{\sqrt{n}}\right)a_n $$ for $n\geq 1$?

We have

  • $(a_n)_n$ is increasing, and $a_n = \prod_{k=1}^n\left(1+\frac{1}{\sqrt{k}}\right)$, from which $$ \ln a_n = \sum_{k=1}^n \ln\left(1+\frac{1}{\sqrt{k}}\right) \operatorname*{\sim}_{n\to\infty} \sum_{k=1}^n \frac{1}{\sqrt{k}} \xrightarrow[n\to\infty]{} \infty $$ so (1) holds.

  • We have $$\frac{a_{n+1}}{a_n} = 1+\frac{1}{\sqrt{n+1} }\xrightarrow[n\to\infty]{} 1$$ so (2) holds.

  • We have $\frac{a_{2n}}{a_n} = \prod_{k=n+1}^{2n}\left(1+\frac{1}{\sqrt{k}}\right)$ from which $$\begin{align} \ln \frac{a_{2n}}{a_n} &= \sum_{k=n+1}^{2n} \ln\left(1+\frac{1}{\sqrt{k}}\right) = \sum_{k=1}^{n} \ln\left(1+\frac{1}{\sqrt{n+k}}\right) \\&\geq n \ln\left(1+\frac{1}{\sqrt{2n}}\right) \operatorname*{\sim}_{n\to\infty} \sqrt{\frac{n}{2}}\xrightarrow[n\to\infty]{} \infty \end{align}$$ so (3) holds.

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    Note: istead of$(1+ \frac{1}{\sqrt{n}})$, I am almost sure that anything of the form $(1+f(n))$ with $f(n) = \omega\left(\frac{1}{n}\right)$ and $f(n)=o(1)$ would work as well. – 2017-02-26
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    Thank a lot for your answer. – 2017-02-26
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    @Schrödinger'sCat You're welcome! – 2017-02-26
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    Nice construction by the way. :-) – 2017-02-26
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We can define $a_n$ recursively by taking $a_1=1$ and $a_{n+1}=a_n\times n^{1/n}$

Clearly we have $\lim\limits_{n\to \infty}\frac{a_{n+1}}{a_n}=\lim\limits_{n\to \infty}n^{1/n}=1$.

On the other hand $\lim\limits_{n\to \infty}\frac{a_{2n}}{a_n}=\lim\limits_{n\to \infty}(n+1)^{1/(n+1)}\dots (2n)^{1/2n}\geq \lim\limits_{n\to \infty}((n)^{1/2n})^n=\lim\limits_{n\to\infty}\sqrt n =\infty$

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    Seems all the constructions are done by recursion. Nice answer, thanks :-) – 2017-02-26
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    @Schrödinger'sCat Well, a recurrence formula is the simplest way to control that both (2) and (3) hold. – 2017-02-26
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Sequence could be set to $$a_n = \left(1+\frac{1}{\log(n)}\right)a_{n_1}$$

You can prove that

$$\frac{a_{2n}}{a_n} = \prod_{i=n}^{2n}\left(1+\frac{1}{\log(i)}\right)\rightarrow \infty$$ while $\frac{a_{n+1}}{a_n} \rightarrow 1$.