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Homeproof of limit of nth root of n

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# proof of limit of nth root of n

In this entry, we present a self-contained, elementary proof of the fact that $\lim_{{n\to\infty}}n^{{1/n}}=1$. We begin by with inductive proofs of two integer inequalities — real numbers will not enter until the very end.

###### Lemma 1.

For all integers $n$ greater than or equal to $5$,

$2^{n}>n^{2}$ |

###### Proof.

We begin with a few easy observations. First, a bit of arithmetic:

$2^{5}=32>25=5^{2}$ |

Second, some algebraic manipulation of the inequality $n>4$:

$\displaystyle n-1$ | $\displaystyle>3$ | ||

$\displaystyle(n-1)^{2}$ | $\displaystyle>9$ | ||

$\displaystyle(n-1)^{2}$ | $\displaystyle>2$ | ||

$\displaystyle n^{2}-2n+1$ | $\displaystyle>2$ | ||

$\displaystyle 2n^{2}$ | $\displaystyle>n^{2}+2n+1$ | ||

$\displaystyle 2n^{2}$ | $\displaystyle>(n+1)^{2}$ |

These observations provide us with the makings of an inductive proof. Suppose that $2^{n}>n^{2}$ for some integer $n\geq 5$. Using the inequality we just showed,

$2^{{n+1}}=2\cdot 2^{n}>2n^{2}>(n+1)^{2}.$ |

Snce $2^{5}>5^{2}$ and $2^{n}>n^{2}$ implies that $2^{{n+1}}>(n+1)^{2}$ when $n\geq 5$ we conclude that $2^{n}>n^{2}$ dor all $n\geq 5$. ∎

###### Lemma 2.

For all integers $n$ greater than or equal to $3$,

$n^{{n+1}}>(n+1)^{n}$ |

###### Proof.

We begin by noting that

$3^{4}=81>64=4^{3}.$ |

Next, we make assume that

$(n-1)^{n}>n^{{(n-1)}}.$ |

for some $n$. Multiplying both sides by $n$:

$n(n-1)^{n}>n^{n}.$ |

Multiplying both sides by $(n+1)^{n}$ and making use of the identity $(n+1)(n-1)=n^{2}-1$,

$n(n^{2}-1)^{n}>n^{n}(n+1)^{n}.$ |

Since $n^{2}>n^{2}-1$, the left-hand side is less than $n^{{2n+1}}$, hence

$n^{{2n+1}}>n^{n}(n+1)^{n}.$ |

Canceling $n^{n}$ from both sides,

$n^{{(n+1)}}>(n+1)^{n}.$ |

Hence, by induction, $n^{{(n+1)}}>(n+1)^{n}$ for all $n\geq 3$. ∎

###### Theorem 1.

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

###### Proof.

Consider the subsequence where $n$ is a power of $2$. We then have

$(2^{m})^{{(1/2^{m})}}=2^{{m/2^{m}}}.$ |

By lemma 1, $m/2^{m}<1/m$ when $m\geq 5$. Hence, $(2^{m})^{{1/2^{m}}}<2^{{1/m}}$. Since $\lim_{{m\to 0}}2^{{1/m}}=1$, and $(2^{m})^{{1/2^{m})}}>1$, we conclude by the squeeze rule that

$\lim_{{m\to 0}}(2^{m})^{{1/2^{m}}}=1.$ |

By lemma 2, the sequence $\{n^{{1/n}}\}$ is decreasing. It is clearly bounded from below by $1$. Above, we exhibited a subsequence which tends towards $1$. Thus it follows that

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

∎

## Mathematics Subject Classification

30-00*no label found*12D99

*no label found*

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