# Find the limit of $\sqrt[n]{n^2 + n}$

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 ?

• 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 $1/n$.) – dan_fulea Dec 2 at 21:32
• 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; $$n^2+n=(1+d_n)^n\ge\binom n3 d_n^3\ .$$ – dan_fulea Dec 2 at 21:38

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!

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 $$L = e^0 = 1$$...

• 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 ? – Dimitris Dimitriadis Dec 2 at 20:56
• @DimitrisDimitriadis not sure what is Bernoulli's Rule... – gt6989b Dec 2 at 21:03

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

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

• Thank you for your answer ! By the way, I wonder why Bernoulli's rule can't be applied to this problem. It works to $\sqrt[n]{n}$ – Dimitris Dimitriadis Dec 2 at 21:38
• 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. – Mark Viola Dec 2 at 22:16

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.

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

Option:

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

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

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.

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 $$n\to\infty$$.