# Calculating $2\sqrt{2}$ to Infinity

• MHB
• maxkor
In summary, the given expression can be written as $2^{1+{1\over2}+{1\over10}+{1\over80}+{1\over880}+\ldots}$, but there is no clear pattern in the exponents of the roots. The expression does not converge according to Mathematica. Attempts have been made to simplify the expression, but the defining recurrence is neither arithmetic nor geometric. This is a highly non-trivial problem for a specific value of n.
maxkor
Calculate $2\sqrt{2\sqrt[5]{2\sqrt[8]{2\sqrt[11]{2 \cdots}}}}$.
I know only that $...=2^{1+{1\over2}+{1\over10}+{1\over80}+{1\over880}+\ldots}$

In order to answer that we will need to know exactly how the exponent are calculated. You have roots of 1, 2, 5, 8, 11. But what next? I don't see any obvious pattern.

14,17,...

Really? I saw that 2+ 3= 5, 5+ 3= 8, 8+ 3= 11 but 1 to 2 does not fit that.

If you want:
calculate $2\ \cdot \ \sqrt{2\sqrt[5]{2\sqrt[8]{2\sqrt[11]{2 \cdots}}}}$.
2 * "that sqrts"

So you need
$$\displaystyle 2 \sqrt{2} \sqrt{ \Pi _{n = 0}^{\infty} 2^{1/(5 + 3n)} }$$

I have no proof but Mathematica says this does not converge.

-Dan

No,
$$\displaystyle 2 \sqrt{2} \sqrt{ \Pi _{n = 0}^{\infty} 2^{1/(5 + 3n)} } \neq 2^{1+\frac12+\frac1{2\cdot5}+\frac1{2\cdot5\cdot8}+\frac1{2\cdot5\cdot8\cdot11}+\cdots}$$

Yes, I was reading that wrong.

Okay, I can't finish it but I can get it started.
$$\displaystyle P = 2\sqrt{2\sqrt[5]{2\sqrt[8]{2\sqrt[11]{2 \cdots}}}}$$

$$\displaystyle P = 2 \cdot 2^{1/2 + 1/2 \cdot 1/5 + 1/2 \cdot 1/5 \cdot 1/8 + 1/2 \cdot 1/5 \cdot 1/8 \cdot 1/11 + \text{ ...}}$$

$$\displaystyle ln(P) = ln(2) + \dfrac{1}{2} ln(2) + \dfrac{1}{2} \cdot \dfrac{1}{5} ln(2) + \dfrac{1}{2} \cdot \dfrac{1}{5} \cdot \dfrac{1}{8} ln(2) + \dfrac{1}{2} \cdot \dfrac{1}{5} \cdot \dfrac{1}{8} \cdot \dfrac{1}{11} ln(2) + \text{ ...}$$

$$\displaystyle ln(P) = ln(2) \left ( \dfrac{1}{2} + \dfrac{1}{2} \cdot \dfrac{1}{5} + \dfrac{1}{2} \cdot \dfrac{1}{5} \cdot \dfrac{1}{8} + \dfrac{1}{2} \cdot \dfrac{1}{5} \cdot \dfrac{1}{8} \cdot \dfrac{1}{11} + \text{ ...} \right )$$

So we have the form
$$\displaystyle ln(P) = ln(2) \left ( \sum_{n = 0}^{ \infty } a_n \right )$$
where $$\displaystyle a_n = \dfrac{ a_{n - 1} }{3n + 2}$$, with $$\displaystyle a_0 = \dfrac{1}{2}$$

The trouble is that the series is neither arithmetic nor geometric. The defining recurrence is $$\displaystyle (3n + 2)a_n - a_{n - 1} = 0$$ and I have never solved one of these before. (I don't even know how to get Mathematica to solve it.) And the, once you get $$\displaystyle a_n$$ you still have to sum it.

This is a highly non-trivial problem for a specific value of n. There may be a way to make it simpler in the limit as n goes to infinity.

-Dan

## 1. What is the value of $2\sqrt{2}$ when calculated to infinity?

The value of $2\sqrt{2}$ when calculated to infinity is an irrational number, meaning it has an infinite number of decimal places and cannot be expressed as a fraction. It is approximately equal to 2.82842712475.

## 2. How do you calculate $2\sqrt{2}$ to infinity?

To calculate $2\sqrt{2}$ to infinity, you can use the formula for finding the limit of a sequence. This involves taking the square root of 2, multiplying it by 2, and then repeating the process infinitely. This will result in an increasingly accurate approximation of the value of $2\sqrt{2}$ to infinity.

## 3. Is $2\sqrt{2}$ to infinity a rational or irrational number?

$2\sqrt{2}$ to infinity is an irrational number. This means it cannot be expressed as a ratio of two integers and has an infinite number of decimal places.

## 4. What is the significance of calculating $2\sqrt{2}$ to infinity?

Calculating $2\sqrt{2}$ to infinity is significant in mathematics as it helps us understand the concept of infinity and irrational numbers. It also has applications in fields such as geometry, physics, and engineering.

## 5. Can $2\sqrt{2}$ ever be calculated to a precise value?

No, $2\sqrt{2}$ cannot be calculated to a precise value as it is an irrational number. However, we can approximate its value to any desired level of accuracy by using advanced mathematical techniques such as calculus.

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