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Can you try to solve it? I tried to do something but I do not know how to continue it: \begin{align} & \lim_{n\to\infty}\sqrt[n]{n^2+3n+1}=\lim_{n\to\infty}(n^2+3n+1)^{1/n} = e^{\lim_{n\to\infty} \frac{1}{n}\ln(n^2+3n+1)} \\[10pt] = {} & e^{\lim_{n\to\infty}\frac{1}{n} \frac{\ln(n^2+3n+1)}{n^2+3n+1} (n^2+3n+1)}. \end{align}

Michael Hardy
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Ghost
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5 Answers5

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For positive values of $n$ we have$$\sqrt[n]{n^2}\lt \sqrt[n]{n^2+3n+1}\le\sqrt[n]{5n^2}$$ Note that $$\lim_{n\to\infty}\sqrt[n]{n}=\lim_{n\to\infty}\sqrt[n]{5}=\color{Red}{1}.$$ Hence by the squeeze we have the answer.

Bumblebee
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  • Why is it like that? – Ghost Nov 07 '16 at 15:51
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    @OvyOvy The first bound is just that $3n+1>0$. The second bound is $3n+1 \leq 4n^2$, which should be clear when $n \geq 1$. So now the problem boils down to just $\lim_{n \to \infty} n^{1/n}$ which is an easier problem. – Ian Nov 07 '16 at 15:55
  • $0\lt 3n+1\le 3n^2+n^2$ – Bumblebee Nov 07 '16 at 16:11
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    I don't know if it's worth noting or not but "5" is just a convenience. if $n > 1$, $ 3n + 1 < 3n^2 + n^2=4n^2$. But if n > 3, say, then $n^2 + 3n + 1 < n^2 + n^2 + n^2/9 = 2\frac 19 n^2 < 3n^2$ and if $n > 4$, then $n^2 + 3n + 1 < n^2 + \frac 34 n^2 + \frac 1{16}n^6 = 1.8125 n^2$ so you could use 2. but that wouldn't be as direct or as simple. – fleablood Nov 07 '16 at 16:59
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$$ \lim_{n\to\infty}\sqrt[n]{n^2+3n+1}\sim \lim_{n\to\infty}\sqrt[n]{n^2}=\lim_{n\rightarrow \infty}n^{2/n}\rightarrow1 $$ If this is not satisfying, you can also note that $$ (n+2)^2=n^2+4n+2 $$ Which is always larger than $n^2+3n+1$ and you can squeeze $$ \lim_{n\to\infty}\sqrt[n]{n^2}\leq\lim_{n\to\infty}\sqrt[n]{n^2+3n+1}\leq\lim_{n\to\infty}\sqrt[n]{(n+2)^2}\\ \Rightarrow \lim_{n\to\infty}n^{2/n}\leq\lim_{n\to\infty}\sqrt[n]{n^2+3n+1}\leq\lim_{n\to\infty}{(n+2)^{2/n}}\\ \Rightarrow 1\leq\lim_{n\to\infty}\sqrt[n]{n^2+3n+1}\leq 1 $$

operatorerror
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PRIMER

In THIS ANSWER, I showed using only the limit definition of the exponential function along with Bernoulli's Inequality that the logarithm function satisfies the inequalities

$$\frac{x-1}{x}\le \log(x) \le x-1 \tag 1$$

for $x>0$.

We will now use $(1)$ to show that $\lim_{n\to \infty}\frac1n \log(n)=0$. To do so, we see from $(1)$ that for any $\alpha>0$

$$\begin{align} \alpha \log(x) &=\log(x^\alpha)\\\\ &\le x^\alpha -1\\\\ &<x^\alpha\\\\ \log(x)\le \frac{x^\alpha}{\alpha} \end{align}$$

Hence, we can assert that

$$\frac1n \log(n)\le \frac{n^{\alpha -1}}{\alpha} \tag 2$$

for any $\alpha >0$.

Since $(2)$ is true for any positive $\alpha$, it is certainly true for $0<\alpha <1$. Therefore, for $0<\alpha <1$ we see that $\lim_{n\to \infty}\frac1n \log(n)=0$.

Next, using $\log(n^2+3n+1)\le \log(5n^2)=\log(5)+2\log(n)$, we see that

$$\begin{align} \lim_{n\to \infty}\sqrt[n]{n^2+3n+1}&=\lim_{n\to \infty}e^{\frac1n \log(n^2+3n+1)}\\\\ &=e^{\lim_{n\to \infty}\frac1n \log(n^2+3n+1)}\\\\ &=e^0\\\\ &=1 \end{align}$$

as was to be shown!

Community
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Mark Viola
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Hint: $$ \frac{1}{n} \ln \left( n^2+3n+1 \right) = \frac{1}{n} \ln \left( n \left(n+3+\frac{1}{n} \right) \right) = \frac{1}{n} \ln n + \frac{1}{n} \ln \left( n+3+\frac{1}{n} \right) $$ Write in the last expression: $$ \frac{1}{n} = \frac{1}{n}-\frac{1}{n+3+\frac{1}{n}}+\frac{1}{n+3+\frac{1}{n}}= \frac{1}{n^2} \frac{3n+1}{n+3n+1} +\frac{1}{n+3+\frac{1}{n}}$$

abcdef
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\lim_{n \to \infty}{1 \over n}\ln\pars{n^{2} + 3n + 1} \\[5px] = &\ \lim_{n \to \infty} \ln\pars{\bracks{n + 1}^{\, 2} + 3\bracks{n + 1} + 1 \over n^{2} + 3n + 1}\qquad \pars{~Stolz-Ces\grave{a}ro\ Theorem~} \\[5mm] = &\ \color{#f00}{0} \qquad\implies\qquad \lim_{n \to \infty}\root[n]{n^{2} + 3n + 1} = \exp\pars{\color{#f00}{0}} = \bbx{\ds{1}} \end{align}

Felix Marin
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