Is it correct that $\lim_{n\to\infty} n(a_n^{1/n}-1)=\lim_{n\to\infty}\log a_n$?

Question

Using

$$ \lim_{n\to \infty} = \frac{a^\frac{1}{n}-1}{\frac{1}{n}}=\ln(a) $$ I was trying to solve this $$\lim_{n\to \infty}n\Bigl(1-\sqrt[n]{\ln(n)}\Bigr )=\lim_{n\to \infty}-\biggl(\frac{\ln^\frac{1}{n}(n)-1}{\frac{1}{n}}\biggr)\rightarrow{-\!}\ln(\ln(n))\rightarrow -\infty $$ My teacher said I should be if sure this was legit and if it was to show a proof, but I don't know if this $$ \lim_{n\to \infty} \frac{(a_n)^\frac{1}{n}-1}{\frac{1}{n}}= \lim_{n\to \infty} \ln(a_n) $$ is always true. This $$ \lim_{n\to \infty} \biggl(\frac{1+\sqrt[n]{n}}{2}\biggr)^\frac{n}{\ln(n)} = \sqrt e$$ is achieved using the same trick and it's the correct answer but that doesn't mean the method is correct. I have little experience in proof writing and don't know how to tackle proving it.

Answer 1

Try expanding the $(a_n)^{1/n}$ as a series, like so:

$$ (a_n)^{\frac1n} = 1 +\frac{\ln a_n}{n} + \frac12\left(\frac{\ln a_n} {n}\right)^2 + \dots $$ using the Taylor series of $\exp(x)$ and the fact that $(a_n)^{1/n} = \exp(\ln(a_n) / n)$. Let $b_n = \frac{\ln a_n} n$. If $b_n$ goes to $0$, we can see that the tail of the series, $\sum_{k>1} \frac1{k!}\left(\frac{\ln a_n} {n}\right)^k$, goes to $0$ asymptotically faster than $b_n$ (since each term is $b_n^k$ for some $k>1$). If $b_n$ does not converge to $0$, then observe that $(a_n)^{1/n}$ does not converge to $1$. This gives us the following result:

Theorem

(1) If $\lim\limits_{n\rightarrow\infty}\frac{\ln a_n}n = 0$, then $\lim\limits_{n\rightarrow\infty} n \left((a_n)^{1/n} - 1\right) = \lim\limits_{n\rightarrow\infty} \ln{a_n}$.

(2) If $\liminf\limits_{n\rightarrow\infty}\frac{\ln a_n}n > 0$, then $\lim\limits_{n\rightarrow\infty} n \left((a_n)^{1/n} - 1\right) = \infty$

Proof of (1): $$\lim\limits_{n\rightarrow\infty} n \left((a_n)^{1/n} - 1\right) = \lim\limits_{n\rightarrow\infty} n \left(1 +\frac{\ln a_n}{n} + \frac12\left(\frac{\ln a_n} {n}\right)^2 + \dots - 1\right) = \lim\limits_{n\rightarrow\infty} n \left(\frac{\ln a_n}{n} + \frac12\left(\frac{\ln a_n} {n}\right)^2 + \dots \right) = \lim\limits_{n\rightarrow\infty}\left(\ln a_n + \frac1{2}\frac{\left(\ln a_n\right)^2}n + \dots \right)= \lim\limits_{n\rightarrow\infty} \ln a_n$$ (note the higher order terms all go to 0).

For (2), observe that if $\lim \frac{\ln a_n} n > 0$, then $\lim (a_n)^{1/n} > 1$.

If you want a proof of (1) without Taylor series, try:

$$ \lim\limits_{n\rightarrow\infty} n \left((a_n)^{1/n} - 1\right) = \lim\limits_{n\rightarrow\infty} \frac{n\left((a_n)^{1/n} - 1\right)\left(\frac{\ln a_n}n\right)}{\left(\frac{\ln a_n}n\right)} = \lim\limits_{n\rightarrow\infty} \frac{\left((a_n)^{1/n} - 1\right)\left(\ln a_n\right)}{\left(\frac{\ln a_n}n\right)} = \left(\lim\limits_{n\rightarrow\infty} \ln{a_n}\right)\left(\lim\limits_{n\rightarrow\infty} \frac{(a_n)^{1/n} - 1}{\frac{\ln a_n}n}\right) = \left(\lim\limits_{n\rightarrow\infty} \ln{a_n}\right)\left(\lim\limits_{n\rightarrow\infty} \frac{\exp\left(\frac{\ln a_n}n\right) - 1}{\frac{\ln a_n}n}\right) $$ so you need to show that $\lim\limits_{n\rightarrow\infty} \frac{\exp\left(\frac{\ln a_n}n\right) - 1}{\frac{\ln a_n}n} = 1$. If $\frac{\ln a_n}n\rightarrow 0$, you can substitute $t = \frac{\ln a_n}n$ and have $\lim\limits_{n\rightarrow\infty} \frac{\exp\left(\frac{\ln a_n}n\right) - 1}{\frac{\ln a_n}n} = \lim\limits_{t\rightarrow 0} \frac{\exp(t) - 1}{t} = \lim\limits_{t\rightarrow 0} \frac{\exp(t) - \exp(0)}{t} = \exp'(0) = 1$.

Answer 2

Use the following. $$ \lim_{n\to \infty} \biggl(\frac{1+\sqrt[n]{n}}{2}\biggr)^\frac{n}{\ln(n)}= \lim_{n\to \infty} \biggl(1+\frac{\sqrt[n]{n}-1}{2}\biggr)^{\frac{2}{\sqrt[n]n-1}\frac{n(\sqrt[n]n-1)}{2\ln(n)}} .$$ Now, after using a substitution $\sqrt[n]n-1=t$ we get the answer: $e^{\frac{1}{2}}.$

Answer 3

Hint:

Rewrite $\;n\Bigl(1-\sqrt[n]{\ln(n)}\Bigr )\;$ as

$$-\ln(n)\,\frac{\mathrm e^{\tfrac{\ln(n)}n}-1}{\cfrac{\ln(n)}n}. $$

Answer 4

Write $a=e^b$. Then just by rewriting your limit $$\lim_{n\rightarrow\infty}\frac{a^{\frac{1}{n}}-1}{\frac{1}{n}} = \lim_{n\rightarrow\infty}\frac{e^{\frac{b}{n}}-1}{\frac{1}{n}}$$

You can then use L'Hopital's rule since both the numerator and denominator tend to $0$.

This is then equal to $$\lim_{n\rightarrow\infty}\frac{\frac{-b}{n^2}e^{\frac{b}{n}}}{-\frac{1}{n^2}} = \lim_{n\rightarrow\infty}be^{\frac{b}{n}}=b=\ln(a)$$ since $e^\frac{b}{n}\rightarrow 0$ as $n\rightarrow\infty$.

Replace $a$ by $a_m$ and take a limit in $m$ so $lim_{m\rightarrow\infty}(\frac{a_m^{\frac{1}{n}}-1}{\frac{1}{n}})=lim_{m\rightarrow\infty} ln(a_m)$ and use that log is continuous, so this is $ln(lim_{m\rightarrow\infty}(a_m))$

Answer 5

The limit $$\lim_{n\to\infty} n(a^{1/n}-1)=\log a\tag{1}$$ is one of the standard / well known limits and can also be taken as the definition of logarithm function.

For the current question the above limit formula does not apply because $\log n$ is not constant. The problem is handled easily by using a generalization of $(1)$ $$\lim_{x\to 0}\frac{a^x-1}{x}=\log a\tag{2}$$ and it's special case when $a=e$ $$\lim_{x\to 0}\frac{e^x-1}{x}=1\tag{3}$$ We have $$\lim_{n\to\infty} n(1-\sqrt[n]{\log n}) =-\lim_{n\to\infty}\dfrac{\exp\left(\dfrac{\log\log n} {n} \right) - 1} {\dfrac{\log \log n} {n} } \cdot \log\log n$$ The first factor tends to $1$ and the second factor tends to $\infty$ so that the desired limit is $-\infty $.

To answer your question concerning a general sequence $\{a_n\} $ with $\sqrt[n] {a_n} \to 1$ we can again note that $$ n(a_n^{1/n}-1)=\dfrac{\exp\left (\dfrac{\log a_n} {n} \right) - 1}{\dfrac{\log a_n}{n}} \cdot\log a_n$$ and the first factor tends to $1$ so the limiting behavior of the sequence $b_n=n(a_n^{1/n}-1)$ is exactly the same as that of the sequence $\log a_n$ provided that $a_n^{1/n}\to 1$.

What happens when $a_n^{1/n}$ does not tend to $1$? Well, if it tends to a limit $L$ with $0\leq L\neq 1$ then clearly the desired limit is $\pm\infty $ depending on sign of $L-1$. On the other hand if $a_n^{1/n}$ diverges (to $\infty$) then the desired limit is $\infty$. Finally if $a_n^{1/n}$ oscillates then the sequence $n(a_n^{1/n}-1)$ oscillates infinitely or diverges.

Answer 6

Let's start with a brief manifesto on the meaning of an assertion of the form $\lim A_n=\lim B_n$: We take it to mean that if one limit exists (in the extended sense, allowing limits of $\infty$ and $-\infty$), then so does the other, and when that happens, the two (extended) limits are the same.

With this understanding, what we want to show is that

$$\lim_{n\to\infty}{a_n^{1/n}-1\over1/n}=\lim_{n\to\infty}\ln(a_n)$$

for any arbitrary sequence $\{a_n\}$ of positive real numbers (so that both sides are well defined).

Now if $a_n\to a\in\mathbb{R^+}$, it suffices to show that

$${a_n^{1/n}-a^{1/n}\over1/n}\to0$$

But

$$a_n^{1/n}-a^{1/n}={a_n-a\over a_n^{(n-1)/n}+a_n^{(n-2)/2}a+\cdots+a_n^{1/n}a^{(n-2)/n}+a^{(n-1)/n}}$$

and

$$n(\min(a_n,a))^{(n-1)/n}\le a_n^{(n-1)/n}+a_n^{(n-2)/2}a+\cdots+a_n^{1/n}a^{(n-2)/n}+a^{(n-1)/n}\le n(\max(a_n,a))^{(n-1)/n}$$

so that

$$(1/n)(a_n^{(n-1)/n}+a_n^{(n-2)/2}a+\cdots+a_n^{1/n}a^{(n-2)/n}+a^{(n-1)/n})\to a$$

(by the Squeeze Theorem), and thus

$${a_n^{1/n}-a^{1/n}\over1/n}={a_n-a\over(1/n)(a_n^{(n-1)/n}+a_n^{(n-2)/2}a+\cdots+a_n^{1/n}a^{(n-2)/n}+a^{(n-1)/n})}\to{0\over a}=0$$

(since we're assuming $a\not=0$). To summarize, if $a_n\to a\in\mathbb{R^+}$, then

$$\lim_{n\to\infty}{a_n^{1/n}-1\over1/n}=\lim_{n\to\infty}\left({a_n^{1/n}-a^{1/n}\over1/n}+{a^{1/n}-1\over1/n} \right)=0+\ln a=\ln a$$

To handle the cases $a_n\to\infty$ and $a_n\to0$, note that

$$a_n\ge M\implies {a_n^{1/n}-1\over1/n}\ge{M^{1/n}-1\over1/n}\to\ln(M)$$

and likewise,

$$a_n\le 1/M\implies {a_n^{1/n}-1\over1/n}\le{(1/M)^{1/n}-1\over1/n}\to-\ln(M)$$

Finally, if the sequence $a_n$ (of positive real numbers) has no limit in $\mathbb{R^+}\cup\{0,\infty\}$, then it has (at least) two subsequences converging to different limits in $\mathbb{R^+}\cup\{0,\infty\}$, in which case so do the sequences $(a_n^{1/n}-1)/(1/n)$ and $\ln a_n$, which means that neither of those sequences has a limit in $\mathbb{R}\cup\{\pm\infty\}$. Thus the assertion

$$\lim_{n\to\infty}{a_n^{1/n}-1\over1/n}=\lim_{n\to\infty}\ln a_n$$

is true for all sequences of positive real numbers $\{a_n\}$.