Collection of Maths Problems

Squeezed sequences

Task number: 2867

• Variant 1

  \(\displaystyle \lim_{n\to\infty} \frac{(-1)^n}n\)

• Hint

  First determine \(\displaystyle \lim_{n\to\infty} \frac1n\).

• Resolution

  We use the fact that if \(\displaystyle \lim_{n\to\infty} |a_n| = 0\), then also \(\displaystyle \lim_{n\to\infty} a_n = 0\).

  This fact follows from the squeeze lemma, because

  \(\displaystyle \lim_{n\to\infty} -|a_n| \leq \lim_{n\to\infty} a_n \leq \lim_{n\to\infty} |a_n| \)

  Because \(\displaystyle \lim_{n\to\infty} \frac1n = 0\), the left and right limits are equal to \(0\).

• Result

  The limit of the sequence is \(0\).

• Variant 2

  \(\displaystyle \lim_{n\to\infty} \frac1n \sin n^2 \)

• Hint

  The sine function is bounded.

• Resolution

  We have \(\displaystyle -\frac1n \leq \frac1n \sin n^2 \leq \frac1n. \)

  The left and right sequences have a limit of \(0\). By the squeeze lemma, the sequence in the middle must also converge to \(0\).

• Result

  The limit of the sequence is \(0\).

• Variant 3

  \(\displaystyle \lim_{n\to\infty}\frac{n}{2^n} \)

• Resolution

  We use the fact that \(n^2\leq 2^n\) for all \(n\geq 4\). So we have the estimate \(\displaystyle 0\leq\frac{n}{2^n}\leq \frac{1}{n}. \)

  Now we use the squeeze lemma and we have \(\displaystyle 0\le \lim_{n\to\infty}\frac{n}{2^n} \le \lim_{n\to\infty}\frac{1}{n} =0 \)

• Result

  The limit of the sequence is \(0\).

• Variant 4

  \(\displaystyle \lim_{n\to\infty}\frac{n^k}{n!} \), pro \(k\in \mathbb N\)

• Hint

  Choose \(n_0\) depending on \(k\).

• Resolution

  If \(n\ge 2k+1\), we get

  \(\displaystyle \frac{n^k}{n!}\le \frac{n}{n-k}\frac{n}{n-k+1}…\frac{n}{n-1}\frac{1}{n} \le \frac{2^k}{n} \)

  It follows that \(\displaystyle 0\le \lim_{n\to\infty}\frac{n^k}{n!} \le \lim_{n\to\infty}\frac{2^k}{n} = 2^k\lim_{n\to\infty}\frac{1}{n} = 2^k\cdot 0 =0 \).

• Result

  The limit of the sequence is \(0\).

• Variant 5

  \(\displaystyle \lim_{n\to\infty}\frac{2^n}{n!}. \)

• Resolution

  We first write the \(n\)-th term of the sequence as a product of \(n\) terms and we make an elementary estimate of most of those terms. We have

  \(\displaystyle 0\leq \frac{2^n}{n!}=\frac{2}{1}\cdot\frac{2}{2}\cdot\frac{2}{3}\cdots\frac{2}{n-1}\cdot\frac{2}{n} \leq 2{\cdot} 1\cdot 1 \cdots 1\cdot \frac{2}{n}=\frac{4}{n}. \)

  Now we use the squeeze lemma and we have \(\displaystyle 0\le \lim_{n\to\infty}\frac{2^n}{n!} \le \lim_{n\to\infty}\frac{4}{n} =0 \)

• Result

  The limit of the sequence is \(0\).

• Variant 6

  \(\displaystyle \lim_{n\to\infty}\frac{q^n}{n!} \), pro \(q\in \mathbb R\).

• Resolution

  Considering the even values of \(n\), we have

  \(\displaystyle \frac{q^n}{n!}=\frac{q}{1}\cdot\frac{q}{2}\cdots\frac{q}{n-1}\cdot\frac{q}{n} =\frac{q^2}{1\cdot n}\cdot\frac{q^2}{2\cdot (n-1)}\cdots \frac{q^2}{(\frac n2-1)\cdot \frac n2}\cdot \)

  All the denominators are larger than \(n\), so \(\displaystyle \frac{q^n}{n!}\leq\left(\frac{q^2}n\right)^{\frac n2} \leq\left(\frac12\right)^{\frac n2} \) for \(n\ge2q^2\).

  Now we use the squeeze lemma: we have \(\displaystyle 0\le \lim_{n\to\infty}\frac{q^n}{n!} \le \lim_{n\to\infty}\frac{1}{2^{\frac n2}} =0 \)

• Result

  The limit of the sequence is \(0\).

• Variant 7

  \(\displaystyle \lim_{n\to\infty}\frac{n!}{n^n} \)

• Resolution

  Again we expand the \(n\)-th term of the sequence as a product of \(n\) terms and use the elementary estimate

  \(\displaystyle 0\leq \frac{n!}{n^n}=\frac{1}{n}\cdot\frac{2}{n}\cdot\frac{3}{n}\cdots\frac{n-1}{n}\cdot\frac{n}{n} \leq \frac{1}{n}\cdot 1{\cdot} 1 \cdots 1{\cdot} 1=\frac{1}{n}. \)

  Now we use the squeeze lemma and we have \(\displaystyle 0\le \lim_{n\to\infty}\frac{n!}{n^n} \le \lim_{n\to\infty}\frac{1}{n} =0 \)

• Result

  The limit of the sequence is \(0\).