1

\begin{align}\lim_{n\to\infty}\sqrt[n]{\frac{\ln\left(\left(1+\frac{1}{n^2}\right)^{n^2}\right)}{n^2}}&=\lim_{n\to\infty}\sqrt[n]{\frac{\frac{1}{n}\ln\left(\left(1+\frac{1}{n^2}\right)^{n^2}\right)}{n}}\\ &=\lim_{n\to\infty}\sqrt[n]{\frac{\ln\left(\sqrt[n]{\left(1+\frac{1}{n^2}\right)^{n^2}}\right)}{n}}\\&= \lim_{n\to\infty}\sqrt[n]{\frac{\ln\left(\sqrt[n]{e}\right)}{n}}\\&=\lim_{n\to\infty}\sqrt[n]{\frac{\ln(1)}{n}}\\&=\lim_{n\to\infty} \frac{\sqrt[n]{0}}{\sqrt[n]{n}}= 0\end{align}

Where is the mistake? It should be $1$ according to the computer

Sine of the Time
  • 8,579
anon
  • 727
  • 3
    Why are you selectively applying limits? That is your mistake. – Myungheon Lee Mar 28 '25 at 13:43
  • @MyungheonLee Why can't I do it? Is it because you can only selectively apply the limit if you actually can rewrite the limit into a combination of other limits (like the limit of a sum is the sum of the limits etc)? – anon Mar 28 '25 at 13:49
  • I'll give you one way: It is equivalent to $\sqrt[n]{\ln(1+\frac{1}{n^2})}$ and use $\ln(1+x)\sim x$, it is equivalent to $\sqrt[n]{\frac{1}{n^2}}$. $\lim_{n\to\infty}\sqrt[n]{n^2}=\lim_{n\to\infty}e^{\frac{2\ln n}{n}}\to 1$. – Bowei Tang Mar 28 '25 at 13:53
  • 3
    According to your method, $\lim_{n\to \infty} \frac nn=\lim_{n\to \infty}n \lim {n\to \infty} \frac 1n=\lim{n\to \infty}n \cdot 0 =0$ – Sine of the Time Mar 28 '25 at 13:54
  • Another example by your method: $\frac{\sin\left(\frac{1}{2n\pi}\right)}{\frac{1}{2n\pi}}\to \frac{\sin 0}{\frac{1}{2n\pi}}=0$, but it tends to $1$ as $n\to \infty$. – Bowei Tang Mar 28 '25 at 14:00
  • @SineoftheTime So what are the only situations where you can split up a limit into multiple limits which you can take separately? Is it really only for sums i.e. $\lim (a_n + b_n) = \lim a_n + \lim b_n$? – anon Mar 28 '25 at 14:29
  • @anon are you asking when the limit of the product is the product of the limits? – Sine of the Time Mar 28 '25 at 15:51
  • @SineoftheTime yes – anon Mar 29 '25 at 08:34

5 Answers5

2

As noticed in the comments you are applying limit in two different steps and this is not allowed in general, refer also to:

In this case we can use for example that:

and since

$$\frac{a_{n+1}}{a_n}=\frac{\ln\left(\left(1+\frac{1}{(n+1)^2}\right)^{(n+1)^2}\right)}{(n+1)^2}\frac{n^2}{\ln\left(\left(1+\frac{1}{n^2}\right)^{n^2}\right)}\to 1 \implies \sqrt[n]{a_n} \to 1$$

user
  • 162,563
0

$$\sqrt[n]{\frac{\ln\left(\sqrt[n]{\left(1+\frac{1}{n^2}\right)^{n^2}}\right)}{n}}\ne\sqrt[n]{\frac{\ln\left(\sqrt[n]{e}\right)}{n}}$$ when $n\to\infty$.

$$\sqrt[n]{\frac{\ln\left(\sqrt[n]{\left(1+\frac{1}{n^2}\right)^{n^2}}\right)}{n}}=\sqrt[n]{\frac{\frac{1}{n}\ln\left({\left(1+\frac{1}{n^2}\right)^{n^2}}\right)}{n}}$$ which tends to $$\sqrt[n]{\frac{\frac{1}{n}}{n}}$$ as $n\to\infty$.

So the answer is indeed $1$.

To evaluate the last limit, use the fact that $n^{\frac{1}{n}}\ge1$ for $n\ge1$ and now let $n^{\frac{1}{n}}=1+r$ where $r\ge0$ and apply sandwich theorem to the $\binom{n}{2}$ term after expanding it.

The expression $$\sqrt[n]{\left(1+\frac{1}{n^2}\right)^{n^2}}$$is not equal to $\sqrt[n]{e}$ for any finite $n$.

The $n$-th root $\sqrt[n]{\cdot}$ is dependent on the same $n$ that is going to infinity.

Blue Cat Blues
  • 10,058
  • Why is it $\neq$? Why can't I just apply the limit inside of the $\ln$ first and then proceed as I've shown? – anon Mar 28 '25 at 13:53
  • 1
    @anon See the comment above by Sine of the Time. – Bowei Tang Mar 28 '25 at 13:57
  • @lurk I've seen, but then, what you did is not right i.e. $$\sqrt[n]{\frac{\frac{1}{n}\ln\left({\left(1+\frac{1}{n^2}\right)^{n^2}}\right)}{n}}$$ which tends to $$\sqrt[n]{\frac{\frac{1}{n}}{n}}$$ as $n\to\infty$. Because you first take the limit of the $\ln$ and then evaluate the other stuff – anon Mar 28 '25 at 14:31
  • @lurk if we can't do $\lim \frac{n}{n} = \lim n \cdot \lim \frac{1}{n} = \lim n \cdot 0 = 0$, then why should we be able to do $\lim \left( \frac{1}{n} \ln\left({\left(1+\frac{1}{n^2}\right)^{n^2}}\right) \right)$ $= \lim \frac{1}{n} \cdot \lim \ln\left({\left(1+\frac{1}{n^2}\right)^{n^2}}\right) $= $\lim \frac{1}{n} \cdot \ln (e) = \lim \frac{1}{n}$ – anon Mar 28 '25 at 14:36
  • @anon (maybe I was wrong) Actually I used an equivalent infinitesimal, $\ln(1+\frac{1}{n^2})\sim \frac{1}{n^2}$, I did not evaluate it as a number. I did it like the following: $${\frac{\ln\left(\left(1+\frac{1}{n^2}\right)^{n^2}\right)}{n^2}}=\frac{n^2\ln\left(1+\frac{1}{n^2}\right)}{n^2}=\ln\left(1+\frac{1}{n^2}\right)\sim \frac{1}{n^2}$$ This change does not affect the limit. – Bowei Tang Mar 28 '25 at 14:37
  • @anon Again, I did not evaluate the limit separately or evaluate it as a number in the middle step. [Just like the two examples given in the comment above] – Bowei Tang Mar 28 '25 at 14:43
0

You could also use Squeeze Theorem .

Since $\;\sqrt e<2\leqslant\!\left(1+\dfrac1{n^2}\right)^{\!n^2}\!\!\!\!<e\;$ for any $\;n\in\Bbb N\;,$
it follows that

$\dfrac1{\sqrt[n]{2n^2}}<\sqrt[n]{\dfrac{\ln\!\left(1+\frac{1}{n^2}\right)^{\!n^2}}{n^2}}<\dfrac1{\sqrt[n]{n^2}}\leqslant1\;\;$ for any $\;n\in\Bbb N\,.$

Moreover, $\;\lim\limits_{n\to\infty}\dfrac1{\sqrt[n]{2n^2}}=\lim\limits_{n\to\infty}\dfrac1{2^{\frac1n}e^{\frac{2\ln n}n}}=\dfrac1{2^0e^0}=1\;,$

hence, by applying Squeeze Theorem, it follows that

$\lim\limits_{n\to\infty}\sqrt[n]{\dfrac{\ln\!\left(1+\frac{1}{n^2}\right)^{\!n^2}}{n^2}}=1\,.$

Angelo
  • 13,952
0

Te error is that there are competing speeds of convergence of sequences similar to what one has with the sequence $(1+\frac{1}{n})^n$. In this case, $\log\left(\Big(1+\frac1{n^2}\Big)^{n^2}\right)\xrightarrow{n\rightarrow\infty}1$ faster that $\frac{1}{n^2}\xrightarrow{n\rightarrow\infty}0$. this gets amplified by taking $n$-th root. Indeed, $$\left(\log\Big(\Big(1+\frac1{n^2}\Big)^{n^2}\Big)\right)^{1/n}=n^{2/n}\left(\log\Big(1+\frac1{n^2}\Big)\right)^{1/n}$$ Notice that $$\frac{1}{(2n^2)^{1/n}}\leq \left(\frac{1}{1+n^2}\right)^{1/n}<\Big(\log\big(1+\frac1{n^2}\big)\Big)^{1/n}<\frac1{n^{2/n}}$$ for $\frac{x}{1+x}<\log(1+x)\leq x$ whenever $1+x>0$ This means that $$\left(\log\Big(\Big(1+\frac1{n^2}\Big)^{n^2}\Big)\right)^{1/n}\xrightarrow{n\rightarrow\infty}1$$ and so, $$ \sqrt[n]{\frac{\ln\left(\left(1+\frac{1}{n^2}\right)^{n^2}\right)}{n^2}}\xrightarrow{n\rightarrow\infty}1$$

Furthermore, we have that $$ \sqrt[n]{\frac{\ln\left(\left(1+\tfrac{1}{n^s}\right)^{n^q}\right)}{n^p}}\xrightarrow{n\rightarrow\infty}1$$ for any $p,q\in\mathbb{R}$ and $s>0$.

Mittens
  • 46,352
0

Your mistake was made at the step

$$\lim_{n\to\infty}\sqrt[n]{\frac{\ln\left(\sqrt[n]{e}\right)}{n}}=\lim_{n\to\infty}\sqrt[n]{\frac{\ln(1)}{n}}$$

It's true that $\lim_{n\to\infty} \sqrt[n]{e} = 1$, but you can't just selectively take the limit of that part independently from the other $n$'s in the expression.

A correct solution is:

$$\lim_{n\to\infty}\sqrt[n]{\frac{\ln\left(\sqrt[n]{e}\right)}{n}}$$ $$= \lim_{n\to\infty}\sqrt[n]{\frac{1/n}{n}}$$ $$= \lim_{n\to\infty} n^{-2/n}$$ $$= \lim_{n\to\infty} \exp\left(\frac{-2 \ln(n)}{n}\right)$$ $$= \exp(0)$$ $$= 1$$

Dan
  • 18,262