Exercise 2.7.10

Question

[Infinite Products]
  Review Exercise 2.4.10 about infinite products and then answer the following questions:
  \begin{enumerate}[label=(\alph*)] 
  \item Does $\frac{2}{1} \cdot \frac{3}{2} \cdot \frac{5}{4} \cdot \frac{9}{8} \cdot \frac{17}{16} \cdots$ converge?
  \item The infinite product $\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdot \frac{7}{8} \cdot \frac{9}{10} \cdots$ certainly converges. (Why?) Does it converge to zero?
  \item In 1655, John Wallis famously derived the formula
    $$
    \left(\frac{2 \cdot 2}{1 \cdot 3}\right)\left(\frac{4 \cdot 4}{3 \cdot 5}\right)\left(\frac{6 \cdot 6}{5 \cdot 7}\right)\left(\frac{8 \cdot 8}{7 \cdot 9}\right) \cdots=\frac{\pi}{2}
    $$
    Show that the left side of this identity at least converges to something. (A complete proof of this result is taken up in Section 8.3.)
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item Rewriting the terms as $a_n = (1 + 1/n)$ and using the result from 2.4.10 implies the product diverges since $\sum 1/n$ diverges.
  \item Converges by the monotone convergence theorem since the partial products are decreasing and greater then zero. To show that the product converges to zero, the key insight is to rewrite each term $a_n = (2n -1) / (2n) = 1 / (2n/(2n - 1)) = 1/b_n$, where $b_n = 2n / (2n -1)$. Then the partial products
$$
    p_n = \prod^n_{i = 1} a_n = \frac{1}{\prod^n_{i = 1} b_n}
$$
But
$$\prod^n_{i=1}b_n = \prod^n_{i=1} \left(1 + \frac{1}{2n - 1}\right)$$
and since $\sum 1/(2n-1)$ diverges by comparison against a multiple of the harmonic series, $\prod b_n$ diverges. Thus, to show there exists some $N$ so that $n \geq N$ implies $p_n < \epsilon$, simply take $N$ large enough so that $\prod^n_{i = 1} b_n > 1/\epsilon$.

\item In compontent form we have
    $$
    a_n
    = \frac{(2n)^2}{(2n-1)(2n+1)}
    = \frac{(2n)^2}{(2n)^2 - 1}
    = 1 + \frac{(2n)^2 - ((2n)^2 - 1)}{(2n)^2 - 1}
    = 1 + \frac{1}{(2n)^2 - 1}
    $$
    And since $\sum 1/((2n)^2 - 1)$ converges by a comparison test with $1/n^2$, exercise 2.4.10 implies
    $$\prod_{n=1}^\infty \left(1 + \frac{1}{(2n)^2 - 1}\right)$$
    also converges.
   \end{enumerate}

Exercise 2.7.10

Answers

Comments

Exercise 2.7.10

Answers

Comments

Add answer