Question

find the limit of the sequence
\[\sqrt{2}, \sqrt{2\sqrt{2}},\sqrt{2\sqrt{2\sqrt{2}}},\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2}}}}}...\]
why is it 2?

Answer 1

Geometric series \[\sum_{n=1}^{\infty} \frac{ 1 }{ 2^n }\]

Answer 2

By writing the sequence using exponents instead of radicals, and simplifying, you get\[\left\{ 2^\frac{ 1 }{ 2 }, 2^\frac{ 3 }{ 4 }, 2^\frac{ 7 }{ 8 }, 2^\frac{ 15 }{ 16 }, ... \right\}\]So you are looking for \[\lim_{n \rightarrow \infty}2^{\frac{ 2^n-1}{2^n}}\]which is 2. Make sense?

Answer 3

how did you change the radicals to exponents starting with 2^3/4?

Answer 4

\[\sqrt{2}=2^\frac{ 1 }{ 2 }\]\[\sqrt{2\sqrt{2}}= 2^\frac{ 1 }{ 2 }\times 2^\frac{ 1 }{ 4 } = 2^\frac{ 3 }{ 4 }\]\[\sqrt{2\sqrt{2\sqrt{2}}} = 2^\frac{ 1 }{ 2 }\times 2^\frac{ 1 }{ 4 }\times 2^\frac{ 1 }{ 8 } = 2^\frac{ 7 }{ 8 }\]and so on.

Answer 5

Very nice @ospreytriple

Answer 6

Or, alternatively,\[\sqrt{\sqrt{\sqrt{2}}} = {{2^\frac{ 1 }{ 2 }}^\frac{ 1 }{ 2 }}^\frac{ 1 }{ 2 } = 2^\frac{ 1 }{ 8 }\]

Answer 7

Thanks @iambatman

Answer 8

how did you get 2^1/2 x 2^1/4 = 2^3/4?
Sorry I'm new with this..

Answer 9

When you multiply exponents you add them

Answer 10

Ok. When you multiply powers with the same base, you retain the base and add the exponents. For example,\[x^2\times x^3 = x^5\]

Answer 11

\(\large \begin{align} \color{black}{\sqrt{2}, \sqrt{2\sqrt{2}},\sqrt{2\sqrt{2\sqrt{2}}},\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2}}}}}\cdot \cdot\cdot\cdot \hspace{.33em}\\~\\

\implies 2^{\frac{1}{2}}, \sqrt{2\cdot 2^{\frac{1}{2}}},\sqrt{2\sqrt{2\cdot 2^{\frac{1}{2}}}},\sqrt{2\sqrt{2\sqrt{2\sqrt{2\cdot 2^{\frac{1}{2}}}}}}\cdot\cdot\cdot\cdot \hspace{.33em}\\~\\
\implies 2^{\frac{1}{2}}, \sqrt{ 2^{\frac{3}{2}}},\sqrt{2\sqrt{ 2^{\frac{3}{2}}}},\sqrt{2\sqrt{2\sqrt{2\sqrt{ 2^{\frac{3}{2}}}}}}\cdot\cdot\cdot\cdot \hspace{.33em}\\~\\
\implies 2^{\frac{1}{2}}, 2^{\frac{3}{4}},\sqrt{2\cdot 2^{\frac{3}{4}}},\sqrt{2\sqrt{2\sqrt{2\cdot 2^{\frac{3}{4}}}}}\cdot\cdot\cdot\cdot \hspace{.33em}\\~\\
\implies 2^{\frac{1}{2}}, 2^{\frac{3}{4}},\sqrt{ 2^{\frac{7}{4}}},\sqrt{2\sqrt{2\sqrt{ 2^{\frac{7}{4}}}}}\cdot\cdot\cdot\cdot \hspace{.33em}\\~\\
\implies 2^{\frac{1}{2}}, 2^{\frac{3}{4}},2^{\frac{7}{8}},\sqrt{2\sqrt{2\cdot 2^{\frac{7}{8}}}}\cdot\cdot\cdot\cdot \hspace{.33em}\\~\\
\implies 2^{\frac{1}{2}}, 2^{\frac{3}{4}},2^{\frac{7}{8}},\sqrt{2\sqrt{ 2^{\frac{15}{8}}}}\cdot\cdot\cdot\cdot \hspace{.33em}\\~\\
\implies 2^{\frac{1}{2}}, 2^{\frac{3}{4}},2^{\frac{7}{8}},\sqrt{2\cdot 2^{\frac{15}{16}}}\cdot\cdot\cdot\cdot\hspace{.33em}\\~\\
\implies 2^{\frac{1}{2}}, 2^{\frac{3}{4}},2^{\frac{7}{8}},\sqrt{ 2^{\frac{31}{16}}}\cdot\cdot\cdot\cdot \hspace{.33em}\\~\\
\implies 2^{\frac{1}{2}}, 2^{\frac{3}{4}},2^{\frac{7}{8}}, 2^{\frac{31}{32}}\cdot\cdot\cdot\cdot \hspace{.33em}\\~\\
\Large \implies 2^{\frac{2^{n-1}}{2^n}} \hspace{.33em}\\~\\
\Large \lim_{n\to\infty} 2^{\frac{2^{n-1}}{2^n}} \hspace{.33em}\\~\\
= \Large \lim_{n\to\infty} 2^{\frac{2^{n}}{2^n\cdot 2}} \hspace{.33em}\\~\\
= \Large \lim_{n\to\infty} 2^{\frac{1}{ 2}} \hspace{.33em}\\~\\
=\sqrt 2 \hspace{.33em}\\~\\
}\end{align}\)

Answer 12

So we have \[2^\frac{ 1 }{ 2 }\times 2^\frac{ 1 }{ 4 } = 2^{\frac{ 1 }{ 2 }+\frac{ 1 }{ 4 }} = 2^\frac{ 3 }{ 4 }\]

Answer 13

oh i see! thanks lots!!!! @ospreytriple

Answer 14

You're welcome.

Answer 15

there is a typo in my soln

Answer 16

@mathmath333 , you've got an error in there somewhere near the end.

Answer 17

But I think we've got it figured out. Thanks.

Answer 18

\(\LARGE \begin{align} \color{black}{\sqrt{2}, \sqrt{2\sqrt{2}},\sqrt{2\sqrt{2\sqrt{2}}},\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2}}}}}\cdot \cdot\cdot\cdot \hspace{.33em}\\~\\

\implies 2^{\frac{1}{2}}, \sqrt{2\cdot 2^{\frac{1}{2}}},\sqrt{2\sqrt{2\cdot 2^{\frac{1}{2}}}},\sqrt{2\sqrt{2\sqrt{2\sqrt{2\cdot 2^{\frac{1}{2}}}}}}\cdot\cdot\cdot\cdot \hspace{.33em}\\~\\
\implies 2^{\frac{1}{2}}, \sqrt{ 2^{\frac{3}{2}}},\sqrt{2\sqrt{ 2^{\frac{3}{2}}}},\sqrt{2\sqrt{2\sqrt{2\sqrt{ 2^{\frac{3}{2}}}}}}\cdot\cdot\cdot\cdot \hspace{.33em}\\~\\
\implies 2^{\frac{1}{2}}, 2^{\frac{3}{4}},\sqrt{2\cdot 2^{\frac{3}{4}}},\sqrt{2\sqrt{2\sqrt{2\cdot 2^{\frac{3}{4}}}}}\cdot\cdot\cdot\cdot \hspace{.33em}\\~\\
\implies 2^{\frac{1}{2}}, 2^{\frac{3}{4}},\sqrt{ 2^{\frac{7}{4}}},\sqrt{2\sqrt{2\sqrt{ 2^{\frac{7}{4}}}}}\cdot\cdot\cdot\cdot \hspace{.33em}\\~\\
\implies 2^{\frac{1}{2}}, 2^{\frac{3}{4}},2^{\frac{7}{8}},\sqrt{2\sqrt{2\cdot 2^{\frac{7}{8}}}}\cdot\cdot\cdot\cdot \hspace{.33em}\\~\\
\implies 2^{\frac{1}{2}}, 2^{\frac{3}{4}},2^{\frac{7}{8}},\sqrt{2\sqrt{ 2^{\frac{15}{8}}}}\cdot\cdot\cdot\cdot \hspace{.33em}\\~\\
\implies 2^{\frac{1}{2}}, 2^{\frac{3}{4}},2^{\frac{7}{8}},\sqrt{2\cdot 2^{\frac{15}{16}}}\cdot\cdot\cdot\cdot\hspace{.33em}\\~\\
\implies 2^{\frac{1}{2}}, 2^{\frac{3}{4}},2^{\frac{7}{8}},\sqrt{ 2^{\frac{31}{16}}}\cdot\cdot\cdot\cdot \hspace{.33em}\\~\\
\implies 2^{\frac{1}{2}}, 2^{\frac{3}{4}},2^{\frac{7}{8}}, 2^{\frac{31}{32}}\cdot\cdot\cdot\cdot \hspace{.33em}\\~\\
\implies 2^{\frac{2^{n}-1}{2^n}} \hspace{.33em}\\~\\
\Large \lim_{n\to\infty} 2^{\frac{2^{n}-1}{2^n}} \hspace{.33em}\\~\\
= \Large \lim_{n\to\infty} 2^{1-\frac{1}{2^n}} \hspace{.33em}\\~\\
= 2^{1-\frac{1}{2^\infty }} \hspace{.33em}\\~\\
= 2 \hspace{.33em}\\~\\
}\end{align}\)

Answer 19

That's it. Good job.

Answer 20

the latex is CRAZY !

Answer 21

@esam2 i think u missed the 4th term in the question

Answer 22

yeah, you're right. There were too many square roots, i lost count. lol @mathmath333
thanks for your help too!

Answer 23

Quick question: how did you guys get the formula, the part where 2^2n-1/2n?
@mathmath333 @ospreytriple

Answer 24

\(2^n=2,4,8,16,32........\infty\)

\(2^{n}-1=1,3,7,15,31........\infty\)