Square root of increasing Exponents of 2

Question

Question: Evaluate $$2\sqrt{2^3\sqrt{2^4\sqrt{2^5\sqrt{2...}}}}$$

I've been trying to look for a pattern in pulling out parts of the exponents (ex. taking a $2^2$ from the $2^3$ in the first square root). I've also been trying to see if I could so something with sigma inside the square roots, but haven't gotten anywhere with that.

Any Ideas? Any and all help is appreciated!

Using the fact that $\sqrt{ab} = \sqrt{a} \sqrt{b}$, you can probably expand your term as $2 \cdot \sqrt{2^3} \cdot \sqrt{\sqrt{2^4}} \cdot \sqrt{\sqrt{\sqrt{2^5}}} \cdots$. Can you see what this gives you? — Jeroen van der Meer, Nov 30 '21 at 19:06
You may want to take $2^2$ from the second, fourth, sixth,..., square roots. — markvs, Nov 30 '21 at 19:07

TheBestMagician · Accepted Answer · 2021-12-01T21:10:29.687

1

Write it as:

$$2\cdot (2^3)^{1/2}\cdot (2^4)^{1/4}\cdot (2^5)^{1/8}...$$

$$2^{1+\frac{3}{2}+\frac{4}{4}+\frac{5}{8}+...}=2^{1+S}$$

Where $$S=\sum_{n=3}^{\infty}\frac{n}{2^{n-2}}=4\sum_{n=3}^{\infty}\frac{n}{2^n}$$

Note that $$\sum_{n=0}^{\infty}\frac{n}{2^n}=2$$ So $$\sum_{n=3}^{\infty}\frac{n}{2^n}=2-\frac{1}{2}-\frac{2}{4}=1$$

Thus $S=4$ and the value is $32$.

edited Dec 01 '21 at 21:10

answered Nov 30 '21 at 19:06

TheBestMagician

2,864

Would S be 1/4, because the n/2^n sigma is multiplied by 1/4 when n=3, or because of the rearranging to make n=0 the 1/4 can be ignored and the solution above works out? – user978757 Dec 01 '21 at 00:52
1

I don't understand. Clearly $S>\frac{1}{4}$ since the first term of $S$ is $\frac{3}{8}>\frac{1}{4}$! – TheBestMagician Dec 01 '21 at 04:00
Wouldn't it be $$4\sum_{n=3}^{\infty}\frac{n}{2^n}$$, since 2^(n-2) is in the denominator, the 2^-2 portion would be 1/4, and you'd have to multiply by 4? – user978757 Dec 01 '21 at 16:15
1

You're right, I'll fix. – TheBestMagician Dec 01 '21 at 21:10

Square root of increasing Exponents of 2

1 Answers1