Find the following limit: $L=\lim_{n\to \infty}\frac{\left(2\sqrt[n]{n}-1\right)^n}{n^2}$

Question

Find the following limit:

$$L=\lim_{n\to \infty}\frac{\left(2\sqrt[n]{n}-1\right)^n}{n^2}$$

I think use Taylor's expansion give $\left(2\sqrt[n]{n}-1\right)^n$ or there is a workaround, but I do not know.

You are on the good track. But, first extract factor n^(1/n) in the numerator, raise to power and use Taylor for the remaining. — Claude Leibovici, Dec 27 '13 at 13:23

score 4 · Answer 1 · answered Dec 27 '13 at 15:40

Let $b_n=\sqrt[n]{n}-1=n^{1/n}-1$ and $a_n=\frac{\left(2\sqrt[n]{n}-1\right)^n}{n^2}$ so that $$a_n=\frac{\left(2\sqrt[n]{n}-1\right)^n}{n^2}=\frac{\left(2\sqrt[n]{n}-2+1\right)^n}{n^2}=\frac{1}{n^2}\operatorname{e}^{n\ln(1+2b_n)}$$ For $n\to \infty, \;b_n\to 0$ because $b_n=\operatorname{e}^{\frac{\ln n}{n}}-1$ and for $n\to \infty$, $\frac{\ln n}{n}\to 0$. So for $n\to\infty$ we have $\ln(1+2b_n)\sim 2b_n$, and $b_n=n^{1/n}-1=\operatorname{e}^{\frac{\ln n}{n}}-1\sim 1+\frac{\ln n}{n}-1=\frac{\ln n}{n}$ so that $$ n\ln(1+2b_n)\sim n2b_n\sim n2\frac{\ln n}{n}=\ln n^2. $$ Finally for $n\to\infty$ $$ a_n=\frac{1}{n^2}\operatorname{e}^{n\ln(1+2b_n)}\sim \frac{1}{n^2}\operatorname{e}^{\ln n^2}=\frac{n^2}{n^2}=1 $$ that is $$ \lim_{n\to\infty}\frac{\left(2\sqrt[n]{n}-1\right)^n}{n^2}=1 $$

The equivalents used in this answer are not enough to deduce the result. One needs a control on the terms neglected to make sure they translate into the limit 1 at the end. — Did, Jul 30 '14 at 08:27

medicu · Answer 2 · 2017-05-03T15:03:08.033

3

$$ \begin{align} L&=\lim_{n\to \infty}\frac{\left(2\sqrt[n]{n}-1\right)^n}{n^2}=\lim_{n\to \infty}\left(\frac{2\sqrt[n]{n}-1}{n^{2/n}}\right)^n \\ &= \operatorname{e}^{\lim_{n\to \infty}\left(\frac{2\cdot\sqrt[n]{n}-1}{n^{2/n}}-1\right)n}=\operatorname{e}^{-\lim_{n\to \infty}\left(\frac{n^{1/n}-1}{n^{1/n}}\right)^2\cdot n}\\ &= \operatorname{e}^{-\lim_{n\to \infty}\left(\frac{\operatorname{e}^{\frac{\ln n}{n}}-1}{{\frac{\ln n}{n}}}\right)^2\cdot\left(\frac{\ln n}{n}\right)^2\cdot \frac{1}{n^{2/n}}\cdot n}\\ &=\operatorname{e}^{-1\cdot 1\cdot \lim_{n\to \infty}\frac{\ln^2n}{n}}=\operatorname{e}^{0}= 1 \end{align}$$

edited May 03 '17 at 15:03

answered Dec 27 '13 at 13:47

medicu

4,482

Might want to sort out the formatting. Looks an bit wrong – Lost1 Dec 27 '13 at 13:48
1

This result is true! Thanks you! – Iloveyou Dec 27 '13 at 13:57
@alexjo, Thanks you! – medicu Dec 27 '13 at 14:21
@medicu It seem's that some is missing in the proof... – alexjo Dec 27 '13 at 14:36
@alexjo. I checked and I have seen mistakes. We used $\lim_{x\to0}\frac{e^x-1}{x}=1$ – medicu Dec 27 '13 at 14:53
@medicu I see...your answer is wrong.. – alexjo Dec 27 '13 at 15:41
It may be wrong. So something is total exclus.Te please show me what I did wrong. Thank you! – medicu Dec 27 '13 at 15:53
@medicu you missed the log in the exponentiation on the second line and the other calculation seems wrong at the beginning of the 2nd line... – alexjo Dec 27 '13 at 16:07
@medicu did you check your answer? – alexjo Dec 27 '13 at 17:11
@Mrnhan this answer is wrong.. – alexjo Dec 27 '13 at 17:11
@alexjo: I just said this result is right – Iloveyou Dec 27 '13 at 17:50
@alexjo. If $\lim_{x\to \alpha} f(x) = 1$ and $\lim_{x\to \alpha} g(x) =\infty $ then there is "the shortcut" $\lim_{x\to \alpha} (f(x))^{g(x)} = e^{\lim_{x\to \alpha} (f(x)-1)g(x)} $ – medicu Dec 27 '13 at 17:51

Find the following limit: $L=\lim_{n\to \infty}\frac{\left(2\sqrt[n]{n}-1\right)^n}{n^2}$

2 Answers2