Collection of Maths Problems

Limits of n-th roots

Task number: 2871

• Variant 1

  \(\displaystyle \lim_{n\to\infty}\root{n}\of{n} \)

• Hint

  First consider the sequence \(\alpha_n\) given by the rule \( \alpha_n=\root{n}\of{n}-1 \).

• Resolution

  We raise to the \(n\)-th power, and with the help of the binomial theorem we have
  \(\displaystyle n=(1+\alpha_n)^n=\sum_{i=0}^{n}\binom{n}{i}\alpha_n^i\geq \binom{n}{2}\alpha_n^2=\frac{n(n-1)}{2}\alpha_n^2 \).

  It follows that \( 0\leq \alpha_n\leq \frac{2}{n-1}, \) and so the sequence \(\alpha_n\) converges to zero.

• Result

  The limit equals \(1\).

• Variant 2

  \(\displaystyle \lim_{n\to\infty}\root{n}\of{n^2+2^n+3^n}. \)

• Resolution

  Recall that for \(n\geq 4\) it is true that \(n^2\leq 2^n\) and evidently \(2^n\leq 3^n\).

  We estimate: \(\displaystyle 3=\root{n}\of{3^n}\leq \root{n}\of{n^2+2^n+3^n}\leq \root{n}\of{3^n+3^n+3^n}= 3\root{n}\of{3} \).

  Finally we use the relsationship \(\lim\limits_{n\to\infty} \root{n}\of{3} =1\) together with the squeeze lemma.

• Result

  The limit is \(3\).

• Variant 3

  \(\displaystyle \lim_{n\to\infty}\sqrt[n]{a^n+b^n+c^n} \), where \(a,b,c>0\) are given real numbers.

• Resolution

  Without loss of generality let \(a \le b\le c\). Then

  \(\displaystyle \lim_{n\to\infty}\sqrt[n]{a^n+b^n+c^n}\ge \lim_{n\to\infty}\sqrt[n]{c^n}=c. \)

  \(\displaystyle \lim_{n\to\infty}\sqrt[n]{a^n+b^n+c^n}\le \lim_{n\to\infty}\sqrt[n]{3c^n} =c\lim_{n\to\infty}\sqrt[n]{3}=c. \)

• Result

  The limit equals the largest of the numbers \(a\), \(b\) and \(c\).

• Variant 4

  \(\displaystyle \lim_{n\to\infty}\sqrt[n]{n!} \)

• Resolution

  We show that for every \(q\) and \(n\ge q^2\) it is true that \(\sqrt[n]{n!}\ge q\).

  That is, \(\displaystyle \sqrt[n]{n!}\ge q \Longleftrightarrow n!\ge q^n \Longleftrightarrow \frac{q^n}{n!} \le 1\).

  We have already proven this analogous inequality when considering \(\displaystyle \lim_{n\to\infty}\frac{q^n}{n!} \).

• Result

  The sequence has the improper limit \(\infty\).