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 ?  
 A: 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 $L = e^0 = 1$...
A: 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
$$\lim_{n\to\infty}\left(n^2+n\right)^{1/n}=1$$
A: 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!
A: 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. 
A: Option:
$1\le (n^2+n)^{1/n} \le (2n^2)^{1/n} =$
$2^{1/n}(n)^{1/n}(n)^{1/n}.$
A: 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.
A: 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 $n\to\infty$.
