The value of the infinite product ${6^{\frac{1}{2}}} \times {6^{\frac{1}{2}}} \times {6^{\frac{3}{8}}} \times {6^{\frac{1}{4}}} \times \ldots $ is

Question

Prepp Team · Accepted Answer

Calculating the Value of an Infinite Product

The problem asks for the value of an infinite product involving the base 6 raised to a sequence of exponents.

The given infinite product is:

${6^{\frac{1}{2}}} \times {6^{\frac{1}{2}}} \times {6^{\frac{3}{8}}} \times {6^{\frac{1}{4}}} \times \ldots $

Using the property of exponents, $a^m \times a^n = a^{m+n}$, an infinite product of the form $\prod_{n=1}^{\infty} a^{x_n}$ can be written as $a^{\sum_{n=1}^{\infty} x_n}$. In this case, the base is $a=6$, and the exponents are $x_n$. So, the value of the product is $6^S$, where $S$ is the sum of the infinite series of exponents.

The sequence of exponents is:

$a_1 = \frac{1}{2}$
$a_2 = \frac{1}{2}$
$a_3 = \frac{3}{8}$
$a_4 = \frac{1}{4}$
and so on...

We need to find the sum $S = \frac{1}{2} + \frac{1}{2} + \frac{3}{8} + \frac{1}{4} + \ldots$.

Finding the Pattern in the Exponents Sequence

Let's examine the given terms in the sequence of exponents to find a pattern. Let the sequence be denoted by $a_n$.

$a_1 = \frac{1}{2}$

$a_2 = \frac{1}{2}$

$a_3 = \frac{3}{8}$

$a_4 = \frac{1}{4}$

Let's try to find if this sequence follows a linear recurrence relation of the form $a_n = c_1 a_{n-1} + c_2 a_{n-2}$ for $n \ge 3$. We can use the given terms to set up a system of equations for the coefficients $c_1$ and $c_2$.

Using $a_1, a_2, a_3$: For $n=3$, $a_3 = c_1 a_2 + c_2 a_1$. Substituting the values, we get:
$\frac{3}{8} = c_1 \left(\frac{1}{2}\right) + c_2 \left(\frac{1}{2}\right)$
Multiplying the equation by 8 to clear the denominators:
$3 = 4c_1 + 4c_2$
Dividing by 4:
$c_1 + c_2 = \frac{3}{4}$ (Equation 1)
Using $a_2, a_3, a_4$: For $n=4$, $a_4 = c_1 a_3 + c_2 a_2$. Substituting the values, we get:
$\frac{1}{4} = c_1 \left(\frac{3}{8}\right) + c_2 \left(\frac{1}{2}\right)$
Multiplying the equation by 8 to clear the denominators:
$2 = 3c_1 + 4c_2$ (Equation 2)

Solving for the Recurrence Coefficients

Now we solve the system of linear equations for $c_1$ and $c_2$:

$c_1 + c_2 = \frac{3}{4}$

$3c_1 + 4c_2 = 2$

From Equation 1, we can express $c_1$ as $c_1 = \frac{3}{4} - c_2$. Substitute this into Equation 2:

$3\left(\frac{3}{4} - c_2\right) + 4c_2 = 2$

$\frac{9}{4} - 3c_2 + 4c_2 = 2$

$c_2 = 2 - \frac{9}{4} = \frac{8}{4} - \frac{9}{4} = -\frac{1}{4}$

Now substitute the value of $c_2$ back into the expression for $c_1$:

$c_1 = \frac{3}{4} - c_2 = \frac{3}{4} - \left(-\frac{1}{4}\right) = \frac{3}{4} + \frac{1}{4} = \frac{4}{4} = 1$

So, the recurrence relation for the exponents is $a_n = 1 \cdot a_{n-1} - \frac{1}{4} a_{n-2}$, which simplifies to $a_n = a_{n-1} - \frac{1}{4} a_{n-2}$ for $n \ge 3$. The initial conditions are $a_1 = \frac{1}{2}$ and $a_2 = \frac{1}{2}$.

Summing the Infinite Series of Exponents

We need to find the sum $S = \sum_{n=1}^{\infty} a_n = a_1 + a_2 + a_3 + a_4 + \ldots$. The terms $a_n$ are generated by the recurrence $a_n - a_{n-1} + \frac{1}{4} a_{n-2} = 0$ for $n \ge 3$.

Assuming the series converges, we can sum the recurrence relation from $n=3$ to infinity:

$\sum_{n=3}^{\infty} \left(a_n - a_{n-1} + \frac{1}{4} a_{n-2}\right) = \sum_{n=3}^{\infty} 0 = 0$

This can be split into individual sums:

$\sum_{n=3}^{\infty} a_n - \sum_{n=3}^{\infty} a_{n-1} + \frac{1}{4} \sum_{n=3}^{\infty} a_{n-2} = 0$

Let $S = \sum_{n=1}^{\infty} a_n = a_1 + a_2 + a_3 + a_4 + \ldots$. We can express the sums in terms of $S$ and the initial terms:

$\sum_{n=3}^{\infty} a_n = a_3 + a_4 + a_5 + \ldots = S - a_1 - a_2$
$\sum_{n=3}^{\infty} a_{n-1} = a_2 + a_3 + a_4 + \ldots = S - a_1$
$\sum_{n=3}^{\infty} a_{n-2} = a_1 + a_2 + a_3 + \ldots = S$

Substitute these expressions back into the summed recurrence relation:

$(S - a_1 - a_2) - (S - a_1) + \frac{1}{4} S = 0$

$S - a_1 - a_2 - S + a_1 + \frac{1}{4} S = 0$

The $S$ terms cancel out, and the $a_1$ terms cancel out:

$-a_2 + \frac{1}{4} S = 0$

Solving for $S$:

$\frac{1}{4} S = a_2$

$S = 4 a_2$

Using the value of the second term, $a_2 = \frac{1}{2}$, we find the sum of the exponents:

$S = 4 \times \frac{1}{2} = 2$

To confirm the series converges, we look at the characteristic equation of the recurrence: $r^2 - c_1 r - c_2 = 0$, which is $r^2 - 1 r - (-\frac{1}{4}) = 0$, or $r^2 - r + \frac{1}{4} = 0$. This equation factors as $\left(r - \frac{1}{2}\right)^2 = 0$. The roots are $r = \frac{1}{2}$ with multiplicity 2. Since the modulus of the root $|\frac{1}{2}| = \frac{1}{2}$ is less than 1, the terms $a_n$ approach 0 as $n \to \infty$, and the sum $S$ converges to a finite value.

Calculating the Infinite Product Value

The value of the infinite product is $6^S$, where $S=2$.

$6^S = 6^2 = 36$

Final Answer

The value of the infinite product ${6^{\frac{1}{2}}} \times {6^{\frac{1}{2}}} \times {6^{\frac{3}{8}}} \times {6^{\frac{1}{4}}} \times \ldots $ is 36.

Revision Table: Infinite Product Calculation

Concept	Explanation
Infinite Product Property	An infinite product of the form $\prod_{n=1}^\infty a^{x_n}$ is equal to $a^{\sum_{n=1}^\infty x_n}$, provided the sum of exponents converges.
Sum of Exponents Series	The problem reduces to finding the sum of the infinite series of exponents: $\frac{1}{2} + \frac{1}{2} + \frac{3}{8} + \frac{1}{4} + \ldots$.
Linear Recurrence Relation	The sequence of exponents $(a_n)$ follows $a_n = a_{n-1} - \frac{1}{4} a_{n-2}$ for $n \ge 3$, with $a_1 = \frac{1}{2}, a_2 = \frac{1}{2}$.
Summing Recurrence Series	For a convergent series from a homogeneous linear recurrence, summing the recurrence equation term by term can determine the total sum $S$. In this case, $-a_2 + \frac{1}{4} S = 0$.
Calculated Sum of Exponents	From $-a_2 + \frac{1}{4} S = 0$ and $a_2 = \frac{1}{2}$, we get $S = 4a_2 = 4 \times \frac{1}{2} = 2$.
Final Product Value	The infinite product is $6^S = 6^2 = 36$.

Additional Information: Infinite Series and Recurrences

Infinite products are extensions of finite products. An infinite product $\prod_{n=1}^{\infty} c_n$ converges to a non-zero value if and only if the series $\sum_{n=1}^{\infty} (c_n - 1)$ converges. For positive terms $c_n$, it converges if and only if $\sum_{n=1}^{\infty} \ln(c_n)$ converges. In our case, $c_n = 6^{a_n}$, so $\ln(c_n) = a_n \ln(6)$. The sum $\sum a_n \ln(6) = \ln(6) \sum a_n$. Since $\ln(6)$ is a constant, the convergence of the infinite product depends directly on the convergence of the series of exponents $\sum a_n$.

Linear recurrence relations describe sequences where each term is a linear combination of previous terms. The behaviour of these sequences is largely determined by the roots of their characteristic equation. For a second-order recurrence $a_n = c_1 a_{n-1} + c_2 a_{n-2}$, the characteristic equation is $r^2 - c_1 r - c_2 = 0$. If all roots have a magnitude less than 1, the terms $a_n$ approach 0 as $n$ goes to infinity, which is a necessary condition for the series $\sum a_n$ to converge.

The method used to find the sum $S$ by summing the recurrence relation is a valid technique when the series converges. It effectively collapses the infinite series into a simple equation involving the sum $S$ and the initial terms of the sequence.

Prepp Team · Answer

6

Prepp Team · Answer

216

Prepp Team · Answer

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