Products

Exercises · Sums and products · with worked solutions · Lecture notes (PDF) · Chapter 2: Products

Exercise 1

Compute:

\[ {\rm a)}~ 5!, \qquad {\rm b)}~ \frac{7!}{5!}, \qquad {\rm c)}~ \frac{10!}{7!}, \qquad {\rm d)}~ \frac{10!}{7!\;3!} \]

Solution

• a) \(5! = \prod_{j=1}^{5} j = 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 = 120\).

• b) Using the formula \(\frac{n!}{(n-k)!} = \prod_{j=1}^{k} (n-j+1)\) with \(n=7\) and \(k=2\): \(\frac{7!}{5!} = 7 \cdot 6 = 42\).

• c) With \(n=10\) and \(k=3\): \(\frac{10!}{7!} = 10 \cdot 9 \cdot 8 = 720\).

• d) From c): \(\frac{10!}{7!\;3!} = \frac{720}{3!} = \frac{720}{6} = 120\).

Exercise 2

Prove that for every natural number \(n \ge 1\) the product of the first \(n\) positive even numbers is:

\[ \prod_{j=1}^{n} 2\:j = 2^n \; n! \]

and compute its value for \(n=5\).

Solution

Using the product by a constant (with \(c=2\) and \(a_j = j\)) we have:

\[ \prod_{j=1}^{n} 2\:j = 2^n \; \prod_{j=1}^{n} j = 2^n \; n! \]

For \(n=5\): \(\prod_{j=1}^{5} 2\:j = 2 \cdot 4 \cdot 6 \cdot 8 \cdot 10 = 2^5 \cdot 5! = 32 \cdot 120 = 3840\).

Exercise 3

Prove that for every natural number \(n \ge 1\) the product of the first \(n\) odd numbers is:

\[ \prod_{j=1}^{n} (2\:j-1) = \frac{(2n)!}{2^n \; n!} \]

and compute its value for \(n=4\).

Solution

The factors of \((2n)! = \prod_{j=1}^{2n} j\) are the \(n\) even numbers \(2, 4, \dots, 2n\) and the \(n\) odd numbers \(1, 3, \dots, 2n-1\). Rearranging the factors and using Exercise Exercise 2, we have:

\[ (2n)! = \prod_{j=1}^{n} 2\:j \; \prod_{j=1}^{n} (2\:j-1) = 2^n \; n! \; \prod_{j=1}^{n} (2\:j-1) \]

and then

\[ \prod_{j=1}^{n} (2\:j-1) = \frac{(2n)!}{2^n \; n!} \]

For \(n=4\): \(1 \cdot 3 \cdot 5 \cdot 7 = 105\) and indeed \(\frac{8!}{2^4 \cdot 4!} = \frac{40320}{16 \cdot 24} = \frac{40320}{384} = 105\).

Exercise 4

• a) Prove that for every sequence of non-zero real numbers \(a_1, a_2, \dots, a_{n+1}\) we have (telescoping product):

  \[ \prod_{j=1}^{n} \frac{a_{j+1}}{a_j} = \frac{a_{n+1}}{a_1} \]

• b) Using a), compute \(\displaystyle \prod_{j=1}^{n} \frac{j+1}{j}\).

Solution

• a) Using the product of products, the index translation and the decomposition, we have:

  \[\begin{align*} \prod_{j=1}^{n} \frac{a_{j+1}}{a_j} &= \frac{\prod_{j=1}^{n} a_{j+1}}{\prod_{j=1}^{n} a_j} = \frac{\prod_{j=2}^{n+1} a_{j}}{\prod_{j=1}^{n} a_j} = \frac{ \left(\prod_{j=2}^{n} a_{j}\right) \; a_{n+1}}{a_1 \; \prod_{j=2}^{n} a_j} = \frac{a_{n+1}}{a_1} \end{align*}\]

• b) Applying a) with \(a_j = j\):

  \[ \prod_{j=1}^{n} \frac{j+1}{j} = \frac{n+1}{1} = n+1 \]

Exercise 5

Prove that for every natural number \(n \ge 2\) we have:

\[ \prod_{j=2}^{n} \left( 1 - \frac{1}{j^2} \right) = \frac{n+1}{2\:n} \]

and compute its value for \(n=10\).

Solution

Since

\[ 1 - \frac{1}{j^2} = \frac{j^2-1}{j^2} = \frac{(j-1)\:(j+1)}{j^2} = \frac{j-1}{j} \; \frac{j+1}{j} \]

from the product of products we have:

\[ \prod_{j=2}^{n} \left( 1 - \frac{1}{j^2} \right) = \prod_{j=2}^{n} \frac{j-1}{j} \; \prod_{j=2}^{n} \frac{j+1}{j} \]

Both products are telescoping (Exercise Exercise 4, with the index starting from \(2\)): with \(a_j = \frac{1}{j}\) we have \(\frac{a_{j}}{a_{j-1}} = \frac{j-1}{j}\), and with \(a_j = j\) we have \(\frac{a_{j+1}}{a_j} = \frac{j+1}{j}\). Then:

\[ \prod_{j=2}^{n} \frac{j-1}{j} = \frac{1}{2} \cdot \frac{2}{3} \cdots \frac{n-1}{n} = \frac{1}{n}, \qquad \prod_{j=2}^{n} \frac{j+1}{j} = \frac{3}{2} \cdot \frac{4}{3} \cdots \frac{n+1}{n} = \frac{n+1}{2} \]

and therefore

\[ \prod_{j=2}^{n} \left( 1 - \frac{1}{j^2} \right) = \frac{1}{n} \; \frac{n+1}{2} = \frac{n+1}{2\:n} \]

For \(n=10\) we get \(\frac{11}{20}\).

Exercise 6

Using the index translation and the index reflection, prove that:

\[ {\rm a)}~ \prod_{j=3}^{n} (j-2) = (n-2)! \quad (n \ge 3), \qquad {\rm b)}~ \prod_{j=0}^{n-1} (n-j) = n! \quad (n \ge 1) \]

and compute the product in a) for \(n=8\).

Solution

• a) With the index translation by \(m=2\) (setting \(i = j-2\), when \(j\) goes from \(3\) to \(n\), \(i\) goes from \(1\) to \(n-2\)):

  \[ \prod_{j=3}^{n} (j-2) = \prod_{i=1}^{n-2} i = (n-2)! \]

  For \(n=8\): \(\prod_{j=3}^{8} (j-2) = 6! = 720\).

• b) With the index reflection (with \(a_j = j\), so that \(a_{n-j} = n-j\)):

  \[ \prod_{j=0}^{n-1} (n-j) = \prod_{j=0}^{n-1} a_{n-j} = \prod_{j=1}^{n} a_j = \prod_{j=1}^{n} j = n! \]

Exercise 7

Prove that for every natural number \(n \ge 1\) we have:

\[ \prod_{j=1}^{n} 2^j = 2^{\frac{n^2+n}{2}} \]

and compute its value for \(n=4\).

Solution

Since the product of powers with the same base is the power with the sum of the exponents (\(2^a \; 2^b = 2^{a+b}\)), we have:

\[ \prod_{j=1}^{n} 2^j = 2^1 \; 2^2 \; {\rm \dots} \; 2^n = 2^{1+2+{\rm \dots}+n} = 2^{\sum_{j=1}^{n} j} = 2^{\frac{n^2+n}{2}} \]

For \(n=4\): \(2^1 \cdot 2^2 \cdot 2^3 \cdot 2^4 = 2^{10} = 1024\).

Exercise 8

Prove that for every natural number \(n \ge 1\) we have:

\[ 2^{n-1} \le n! \le n^n \]

Solution

We use the fact that if \(0 \le a_j \le b_j\) for every \(j \in \{1, \dots, n\}\), then \(\prod_{j=1}^{n} a_j \le \prod_{j=1}^{n} b_j\) (the inequalities between non-negative numbers can be multiplied member to member).

• Upper bound: since \(j \le n\) for every \(j \in \{1,\dots,n\}\), using the product with constant term we have:

  \[ n! = \prod_{j=1}^{n} j \le \prod_{j=1}^{n} n = n^n \]

• Lower bound: for \(n=1\) we have \(2^0 = 1 = 1!\). For \(n \ge 2\), using the decomposition and since \(j \ge 2\) for every \(j \in \{2,\dots,n\}\), we have:

  \[ n! = 1 \cdot \prod_{j=2}^{n} j \ge \prod_{j=2}^{n} 2 = 2^{n-1} \]

Exercise 9

Prove that \((n+1)! = (n+1)\; n!\) for every natural number \(n \ge 0\), and use it to simplify, for \(n \ge 1\), the expression

\[ \frac{(n+1)! - n!}{(n-1)!} \]

Then compute its value for \(n=6\).

Solution

For \(n=0\) we have \(1! = 1 = 1 \cdot 0!\). For \(n \ge 1\), from the decomposition:

\[ (n+1)! = \prod_{j=1}^{n+1} j = \left(\prod_{j=1}^{n} j\right) \; (n+1) = (n+1) \; n! \]

Then, using it twice (\(n! = n \; (n-1)!\) holds for \(n \ge 1\)):

\[ \frac{(n+1)! - n!}{(n-1)!} = \frac{(n+1)\; n! - n!}{(n-1)!} = \frac{n \; n!}{(n-1)!} = \frac{n \; n \; (n-1)!}{(n-1)!} = n^2 \]

For \(n=6\): \(\frac{7! - 6!}{5!} = \frac{5040 - 720}{120} = \frac{4320}{120} = 36 = 6^2\).