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Suppose that $\lim_{n\rightarrow\infty}\sqrt[n]{a_n}=1$ and $\lim_{n\rightarrow\infty}\frac{b_n}{a_n}=g$ where $g\in (0,\infty)$. We must prove that $\lim_{n\rightarrow\infty} \sqrt[n]{b_n}=1$.

Intuition tells me that this should be done by contradiction. But I can't get it to work. Here is what I came up with so far:

We know that for every $\epsilon>0$, for sufficiently large $n$ we have $\sqrt[n]{a_n}<\epsilon+1$. Suppose that $\lim_{n\rightarrow\infty} \sqrt[n]{b_n}\neq 1$. That means that there exists $\epsilon'$ such that, for sufficiently large $n'$ we have $\sqrt[n]{b_n}>\epsilon' + 1$.

Let's take $\epsilon'$ and $n=\max(n, n')$. Then we have:

$\sqrt[n]{b_n}>\epsilon' + 1$

$\sqrt[n]{a_n}<\epsilon' + 1$

And here, I think, something smart should be done to conclude that, in that case $\lim_{n\rightarrow\infty}\frac{b_n}{a_n}\neq g$. But I just can't see what to do. Any hints?

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    $\begingroup$ I'd argue directly. For large enough $n$, we have $$\frac{g}{2} < \frac{b_n}{a_n} < 2g.$$ $\endgroup$ Dec 7, 2014 at 13:57
  • $\begingroup$ What if $g=+\infty$? $\endgroup$
    – qiubit
    Dec 7, 2014 at 14:05
  • $\begingroup$ Doesn't $g \in (0, \infty)$ mean that $0<g<\infty$? $\endgroup$
    – desos
    Dec 7, 2014 at 14:06
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    $\begingroup$ That is excluded by $g\in (0,\infty)$. If $\frac{b_n}{a_n}\to +\infty$, then it need not hold that $\sqrt[n]{b_n}\to 1$. $\endgroup$ Dec 7, 2014 at 14:07
  • $\begingroup$ Is there a contrarguement to that? $\endgroup$
    – qiubit
    Dec 7, 2014 at 14:10

3 Answers 3

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Assume $\lim_{n \to \infty} (b_n)^{1/n} = M \ne 1$ and assume that the limit of $b_n$ exists:

$|(b_n)^{1/n} - M| < \epsilon$ if $n > N$

Also, $\lim_{n\to 1} (b_n)^{1/n} = b_1 \implies | (b_n)^{1/n} - 1| < \epsilon$

Let $\epsilon = |b_1 - M|/2$

$|b_1 - M|/2 = |b_1 - (b_n)^{1/n} + (b_n)^{1/n} - M|/2 = |-((b_n)^{1/n} - b_1) + ((b_n)^{1/n} - M)|/2 < |(b_n)^{1/n} - b_1|/2 + |(b_n)^{1/n} - M|/2$

$\therefore |(b_n)^{1/n} - M| < |b_1 - M|/2 < |(b_n)^{1/n} - M|/2 < |(b_n)^{1/n} - b_1|/2 + |(b_n)^{1/n} - M|/2 < |b_1 - M|$

A contradiction.

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Let $\varepsilon > 0$. From $\lim_{n \to \infty} b_n/a_n = g \in (0,\infty)$ follows that there exists $n_0 \in \mathbb{N}$ s.t. \begin{align*} n \geq n_0 &\implies \left| \frac{b_n}{a_n} - g \right| < \varepsilon \\ &\iff g - \varepsilon < \frac{b_n}{a_n} < g + \varepsilon \end{align*} We can assume that $g - \varepsilon > 0$. Now \begin{align*} \implies \sqrt[n]{g-\varepsilon} &< \frac{\sqrt[n]{b_n}}{\sqrt[n]{a_n}} < \sqrt[n]{g + \varepsilon} \\ \iff \sqrt[n]{g-\varepsilon}\sqrt[n]{a_n} &< \sqrt[n]{b_n} < \sqrt[n]{g + \varepsilon}\sqrt[n]{a_n} \end{align*}

The left and right sides of the last inequality tend to $1$ as $n \to \infty$, so the result follows.

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  • $\begingroup$ How does $(g - \epsilon)^{1/n}$ tend to $1$ and $n \to \infty$ $\endgroup$
    – Amad27
    Dec 7, 2014 at 14:25
  • $\begingroup$ Because $g > 0$ we can consider epsilons that are less than $g$ so $g - \varepsilon$ is positive and I assumed it's a known result that then $\sqrt[n]{g-\varepsilon}$ tends to 1. $\endgroup$
    – desos
    Dec 7, 2014 at 14:29
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Is this solution correct?

For sufficiently large $n$: $\frac{g}{2} < \frac{b_n}{a_n} < 2g$. If we take the n-th degree root of all these terms we get:

$\sqrt[n]{\frac{g}{2}} < \frac{\sqrt[n]{b_n}}{\sqrt[n]{a_n}}<\sqrt[n]{2g}$. $g$ is finite so $\lim$ of left and right side evaluates to $1$. That means that $\lim \frac{\sqrt[n]{b_n}}{\sqrt[n]{a_n}}=1$. But $\lim \sqrt[n]{a_n}=1$ so in order for that to be true $\lim \sqrt[n]{b_n}$ must be equal to $1$ as well.

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    $\begingroup$ Looks ok to me. $\endgroup$
    – desos
    Dec 7, 2014 at 14:25

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