Collection of Maths Problems

Limits of products

Task number: 2902

• Variant 1

  \(\displaystyle \lim_{n\to\infty}\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)…\left(1-\frac{1}{n^2}\right). \)

• Resolution

  For every \(j\in\{2{,}3,…,n\}\) we can rewrite the corresponding term as \[ 1-\frac{1}{j^2}=\frac{j^2-1}{j^2}=\frac{(j-1)(j+1)}{j^2}. \] So most terms in the product cancel out, and we have \[ \left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)…\left(1-\frac{1}{n^2}\right) =\frac{1{\cdot} 3}{2^2}\cdot\frac{2{\cdot} 4}{3^2}\cdot\frac{3{\cdot} 5}{4^2}\cdots\frac{(n-2)n}{(n-1)^2}\cdot\frac{(n-1)(n+1)}{n^2}=\frac{1}{2}\frac{n+1}{n}. \] Now we can use the arithmetic of limits.

• Result

  The limit is \(\frac{1}{2}\).

• Variant 2

  \(\displaystyle \lim_{n\to\infty}\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdots\frac{2n-1}{2n}. \)

• Resolution

  Let \[ a_n=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdots\frac{2n-1}{2n}\text{ a }b_n=\frac{2}{3}\cdot\frac{4}{5}\cdots\frac{2n}{2n+1}. \] Most terms in the product of these two sequences cancel out, and we have \[ a_n b_n=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdots\frac{2n-1}{2n}\cdot\frac{2n}{2n+1}=\frac{1}{2n+1}. \] Furthermore, for every \(j\in\{1{,}2,…,n\}\) we have \(\frac{2j-1}{2j}\leq \frac{2j}{2j+1}\) and the product of these \(n\) inequalities gives us \(a_n\leq b_n\). So \[ 0\leq a_n\leq \sqrt{a_n b_n}\leq \frac{1}{\sqrt{2n+1}} \] Now we use the squeeze lemma.

• Result

  The limit equals \(0\).

• Variant 3

  \(\displaystyle \lim_{n\to\infty}\sqrt{2}\cdot\sqrt[4]{2}\cdot\sqrt[8]{2}\cdots\sqrt[2^n]{2} \)

• Resolution

  \(\displaystyle \lim_{n\to\infty}\sqrt{2}\cdot\sqrt[4]{2}\cdot\sqrt[8]{2}\cdots\sqrt[2^n]{2}= \lim_{n\to\infty}2^{\frac12}\cdot2^{\frac14}\cdot2^{\frac18}\cdots2^{\frac1{2^n}}= \lim_{n\to\infty}2^{\frac12+\frac14+\frac18+\cdots+\frac1{2^n}}= \lim_{n\to\infty}2^{\frac{1-\frac1{2^n}}{2\left(1-\frac12\right)}}= \)
  \(\displaystyle =2^{1-\lim_{n\to\infty}\frac1{2^n}}= 2^{1-0}=2 \)

• Result

  The limit equals \(2\).

• Variant 4

  \(\displaystyle \lim_{n\to\infty}\left(\frac{2^3-1}{2^3+1}\cdot\frac{3^3-1}{3^3+1}\cdots\frac{n^3-1}{n^3+1}\right) \).

• Result

  \(\frac23\)