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To find the limit I got the $\sqrt[3n]{n^2+n}$

Particularly, $\sqrt[3n]{n^2+n} \ge 1 \rightarrow \sqrt[3n]{n^2+n} = 1 + d_n$ where $d_n\ge 0$.

According to the Bernoulli's rule $\sqrt{n^2+n} = (1+d_n)^n \ge d_n\cdot n \rightarrow d_n \le \frac{\sqrt{n^2+n}}{n}$

The $\frac{\sqrt{n^2+n}}{n} \rightarrow 1$, so $\lim d_n=1 $

So, $\lim\sqrt[n]{n^2+n} = \lim (1+d_n)^3 = \lim(1+3d_n^2+3d_n+d_n^3) =8$

However, $\sqrt[n]{n^2+n}$ tends to $1$. Where is the problem of my solution ? Can you give me a hint of how can I solve it with Bernoulli's rule ?

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  • $\begingroup$ If we use the raw double estimation $1\le n^2+n\le n^9$ for $n\ge 2$, what do we obtain? (Here i assume we know the limit of $n$ to the power $1/n$.) $\endgroup$ – dan_fulea Dec 2 at 21:32
  • $\begingroup$ Oh, the question was somehow different, well, we need slightly more than the simple Bernoulli inequality if we want to adapt the argument from the convergence of $\sqrt[n]n$ to one. For instance; $$n^2+n=(1+d_n)^n\ge\binom n3 d_n^3\ .$$ $\endgroup$ – dan_fulea Dec 2 at 21:38
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The fundamental mistake in your solution is that $d_n\le{\sqrt{n^2+n}\over n}\to1$ does not imply that $\lim_{n\to\infty}d_n=1$, but only that $\lim_{n\to\infty}d_n\le1$, from which all you can say is that $\lim_{n\to\infty}\sqrt[n]{n^2+n}=\lim_{n\to\infty}(1+d_n)^3\le(1+1)^3=8$, which is true but not helpful.

You actually can make your approach work, by correcting another error in it. You went, incorrectly, from $\sqrt[3n]{n^2+n}=1+d_n$ to $\sqrt{n^2+n}=(1+d_n)^n$ instead of $\sqrt[3]{n^2+n}=(1+d_n)^n$. While it's uselessly true that ${\sqrt{n^2+n}\over n}\to1$, it is usefully true that ${\sqrt[3]{n^2+n}\over n}\to0$, because the inequality $\sqrt[3]{n^2+n}=(1+d_n)^n\ge d_n\cdot n$ now says that $0\le d_n\le{\sqrt[3]{n^2+n}\over n}\to0$, so $\lim_{n\to\infty}d_n=0$ and thus

$$\lim_{n\to\infty}\sqrt[n]{n^2+n}=\lim_{n\to\infty}(1+d_n)^3=(1+0)^3=1$$

Remark: I might add, the notion of evaluating the limit of an $n$th root by starting with a root of the $n$th root (in this case the cube root of the $n$th root) is a novel approach, one I for one had not seen before. It's nice to learn something new!

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  • $\begingroup$ Perfect answer to my question. You answered exactly my answer. Thank you. $\endgroup$ – Dimitris Dimitriadis Dec 2 at 21:57
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If your limit exists, say it is equal to $L$. Then $$ \begin{split} \ln L &= \ln \left( \lim_{n \to \infty} \left(n^2+n\right)^{1/n} \right) \\ &= \lim_{n \to \infty} \ln \left( \left(n^2+n\right)^{1/n} \right) \\ &= \lim_{n \to \infty} \frac{\ln \left(n^2+n\right)}{n} \\ &= \lim_{n \to \infty} \left[ \frac{\ln n + \ln (n+1)}{n} \right] \\ &= 0, \end{split} $$ which implies that $L = e^0 = 1$...

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  • $\begingroup$ Thank you for you answer. Could you also give me the solution with Bernoulli's rule ? Or, a hint that my proof is not correct ? $\endgroup$ – Dimitris Dimitriadis Dec 2 at 20:56
  • $\begingroup$ @DimitrisDimitriadis not sure what is Bernoulli's Rule... $\endgroup$ – gt6989b Dec 2 at 21:03
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First, we will show that $\lim_{n\to\infty}n^{1/n}=1$. We define the sequence $x_n$ as

$$x_n=n^{1/n}-1\tag1$$

ASIDE:

Note that $n^{1/n}\ge 1$for $n\ge1$. This is true since if $0\le y\le 1$, then $0\le y^n\le 1$ for all $n\in \mathbb{N}$.


From $(1)$, it is easy to see that $(1+x_n)^n=n$. Then, using the binomial theorem, we see that

$$\begin{align} n&=(1+x_n)^n\\\\ &=\sum_{k=0}^n \binom{n}{k}x_n^k\\\\ &\ge \binom{n}{2}\,x_n^2\\\\ &=\frac{n(n-1)}{2}\,x_n^2 \tag2 \end{align}$$

from which we conclude that $$\begin{align} 0\le x_n \le \sqrt{\frac{2}{n-1}}\tag3 \end{align}$$

Applying the squeeze theorem to $(3)$ reveals

$$\lim_{n\to\infty}n^{1/n}=1\tag4$$

Finally, we write

$$\left(n^2+n\right)^{1/n}=n^{1/n}\left(1+\frac1n\right)^{1/n}\tag5$$

Inasmuch as the limit of the second term on the right-hand side of $(5)$ is not of indeterminate form, rather is of the form $1^1=1$, we conclude from using $(4)$ that

$$\lim_{n\to\infty}\left(n^2+n\right)^{1/n}=1$$

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  • $\begingroup$ Thank you for your answer ! By the way, I wonder why Bernoulli's rule can't be applied to this problem. It works to $\sqrt[n]{n}$ $\endgroup$ – Dimitris Dimitriadis Dec 2 at 21:38
  • $\begingroup$ Well, yes. If you wish to use Bernoulli's Inequality to show$n^{1/n}\to 1$ instead of the binomial theorem, then this approach works. $\endgroup$ – Mark Viola Dec 2 at 22:16
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I suppose the question is: What is wrong with your proof?

You showed $d_n ≤ {(n^2+n)^{1/2} \over n}$. Take a spreadsheet to see the first 20 values and you will see that $d_n$ is actually a lot less than $(n^2+n)^{1/2} \over n$. $d_n ≤ {(n^2+n)^{1/2} \over n}$ doesn't imply that the limit of $d_n$ equals the limit of $(n^2+n)^{1/2} \over n$, it implies that it is less or equal. In this case: A lot less. 0 and not 1.

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  • $\begingroup$ How can one say that $d_n$ is a lot less than this fraction? So, how could I solve it with Bernoulli's rule ? $\endgroup$ – Dimitris Dimitriadis Dec 2 at 21:04
  • $\begingroup$ Coud we say that $ d_n \le \frac{1}{\sqrt{n}} \le \frac{\sqrt{n^2 + n}}{n} $ ? $\endgroup$ – Dimitris Dimitriadis Dec 2 at 21:07
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Option:

$1\le (n^2+n)^{1/n} \le (2n^2)^{1/n} =$

$2^{1/n}(n)^{1/n}(n)^{1/n}.$

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Hint: This is pretty straightforward if you accept that $\sqrt[n]n\to1$. For, $\sqrt[n]{n^2+n}=\sqrt[n]n\cdot \sqrt[n]{n+1}$. Note $\sqrt[n]{n+1}\to1$ easily.

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How to fix your approach

You define $d_n=\sqrt[3n]{n^2+n}-1$. This then implies that $\sqrt[\large\color{#C00}{3}]{n^2+n}=(1+d_n)^n\ge nd_n$ (the $3$ was left out). Thus, $d_n\le\frac{\sqrt[3]{n^2+n}}n=\sqrt[\large3]{\frac{n+1}{n^2}}\to0$.


A Different Bernoulli Approach

Bernoulli's Inequality says $$ \begin{align} n^2+n &\le\left(1+\sqrt{n}\right)^4\tag{expand}\\ &\le\left(1+\frac1{\sqrt{n}}\right)^{4n}\tag{Bernoulli} \end{align} $$ Therefore, $$ \sqrt[\large n]{n^2+n}\le\left(1+\frac1{\sqrt{n}}\right)^4 $$ which tends to $1$ as $n\to\infty$.

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