Calculate approximately the expression $A = 5^{1/2} . 5^{1/4} . 5^{1/8}...$

Question

Calculate approximately the expression $A = 5^{1/2} \cdot 5^{1/4} \cdot 5^{1/8}\cdot\ldots$

My books says to says to do this:

$ 5^{1/4} \cdot 5^{1/8} \cdot 5^{1/16}\cdot\ldots = A $

Then

$ A = \sqrt {5A} \Longrightarrow A^2 = 5A \Longrightarrow A =5 $

Why $ 5^{1/4}\cdot 5^{1/8}\cdot 5^{1/16}\cdot\ldots = A $ ??

Thank you.

Answer 1

Maybe you are asking why $A=\sqrt{5A}$?

Since $$5A=5^1\cdot 5^{1/2}\cdot 5^{1/4}\cdots,$$ we have $$\sqrt{5A}=(5A)^{1/2}=5^{1/2}\cdot 5^{1/4}\cdot 5^{1/8}\cdots=A.$$

Answer 2

Because when you take square roots, you have to divide the exponent by $2$.

Clearly your book wants you to solve this by setting up an equation for the answer and finding the answer. The direct way would be $$ 5^{1/2}\cdot 5^{1/4} \cdots = 5^{1/2+1/4+1/8+\cdots} = 5 ^1 = 5$$

Answer 3

$$5^{1/2} 5^{1/4}\cdot5^{1/8}\cdot5^{1/16}... =5^{\frac12+\frac14+\frac18+\frac1{16}}$$

As the power of $5$ are in infinite Geometric Progression with common ratio $=\frac12<1$

$$\frac12+\frac14+\frac18+\frac1{16}=\frac12\cdot\frac1{1-\frac12}=1$$

Answer 4

You are considering the infinite product $$A = \prod_{k=1}^\infty 5^{1/2^k} = \lim_{n\to \infty}\prod_{k=1}^n 5^{1/2^k}.$$ In order to calculate the limit, we have to be sure that the limit exists. This is the case in your example, as we can see by taking the logarithm: $$\log(\prod_{k=1}^n 5^{1/2^k}) = \sum_{k=1}^n\log(5^{1/2^k}) = \log(5) \sum_{k=1}^{n}\frac{1}{2^k} \text{converges for }n\to \infty.$$

Considering partial product, we find that $$\left(\prod_{k=1}^n 5^{1/2^k} \right)^2 = \prod_{k=1}^n 5^{2/2^k} = \prod_{k=1}^n 5^{1/2^{k-1}} = 5\prod_{k=1}^{n-1} 5^{1/2^k}.$$ Taking limits $n\to\infty$ on both sides gives $A^2 = 5A$. Note that this does not show $A = 5$ immediately, as $A=0$ would be another solution to the equation. But all powers of $5$ are $>1$, therefore $A>1$ and hence $A=5$ is the only remaining solution.

Why is it important that the limit exists?

We have concluded $A^2=5A$ by taking limits on both sides of an equation. But the non-existence of both limits is also a possible result in general. Take for example: $$\sum_{n=0}^\infty 2^n = 1+\sum_{n=1}^\infty 2^n = 1+2\sum_{n=0}^\infty 2^n. \\ \not \Rightarrow \sum_{n=0}^\infty 2^n = -1.$$Apparently, something went wrong. The problem is exactly that the limits on both sides of the equation do not exist. We have to be very careful when dealing with infinity!