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I didn't know how to calculate this:

$$\sqrt{3\sqrt{5\sqrt{3\sqrt{5\cdots}}}}$$

Please help me. Thanks.

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6 Answers 6

up vote 19 down vote accepted

$$\sqrt{3\sqrt{5\sqrt{3\sqrt{5\cdots}}}}=3^{\frac{1}{2}}\cdot 3^{\frac{1}{8}}\cdots 5^{\frac{1}{4}}\cdot 5^{\frac{1}{16}}\cdots=3^{\frac{1}{2}+\frac{1}{8}+\cdots}\cdot 5^{\frac{1}{4}+\frac{1}{16}+\cdots}=3^{\frac{\frac{1}{2}}{1-\frac{1}{4}}}\cdot 5^{\frac{\frac{1}{4}}{1-\frac{1}{4}}}$$ $$=3^{\frac{\frac{1}{2}}{\frac{3}{4}}}\cdot 5^{\frac{\frac{1}{4}}{\frac{3}{4}}}=\sqrt[3]{45}$$

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1  
Oh!! I am happy. This looks so nice if it was left just by writing, thanky very much –  user100991 Jan 7 at 11:53

$$\text{Let }A=\sqrt{3\sqrt{5\sqrt{3\sqrt{5\cdots}}}}$$

Squaring we get $$A^2=3\sqrt{5\sqrt{3\sqrt{5\cdots}}}$$

Again, squaring we get $$A^4=3^2\cdot 5\sqrt{3\sqrt{5\cdots}}=45A$$

Clearly, $A>0$

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thanks sir, for your answer –  user100991 Jan 7 at 11:52
2  
If is ok then you should mark this as answer. So that we know there is an answer that worked for you. Thanks. –  Umberto Jan 7 at 11:56
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@user100991, my pleasure. Please find the proof of converge here(math.stackexchange.com/questions/589288/…) –  lab bhattacharjee Jan 7 at 15:00
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Very nice solution –  qwr Jul 25 at 2:09

$$\begin{eqnarray} A&=&\ \ {\sqrt{3\sqrt{5\sqrt{3\sqrt{5\cdots}}}}} \\ \,&=&\ \ \sqrt{\sqrt{3^2\cdot5\sqrt{\sqrt{3^2\cdot5\cdots}}}}\\ \,&=&\ \ \sqrt[4]{45\sqrt[4]{45\sqrt[4]{45\sqrt[4]{45\cdots}}}}\\ \,&=&\ \ \sqrt[4]{45}\cdot \sqrt[16]{45}\cdot \sqrt[64]{45}\cdots\\ \,&=&\ \ 45^{\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\dots}\\ \end{eqnarray}$$

Since $\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\dots = \frac{1}{4}\cdot \frac{1}{1-\frac{1}{4}}=\frac{1}{3}$

$$A=\sqrt[3]{45}$$

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You have the sequence $a_{n+1}=\sqrt{3\sqrt{5a_n}}$ with $a_0=1$, before calculating its limit, you must prove that it is, in fact, a number, to do so, define $f: [1,\infty)\rightarrow [1,\infty)$ such that $x\mapsto \sqrt{3\sqrt{5x}}$, now you must prove that this function is a contraction, for that, take the fact that $|\sqrt{3\sqrt{5x}}-\sqrt{3\sqrt{5y}}|\leq |f'(c)| |x-y|$ (mean-value theorem), also $|f'(c)|=\frac{\sqrt{3}5^{\frac{1}{4}}}{4c^{\frac{3}{4}}}<1$ as $1\leq c$, we can assume that because $a_0=1$ and $a_n$ is an increasing sequence, thus $f$ is a contraction and by the Banach Fixed Point Theorem $f$ has a fixed point $f(x)=x$ such that $a_n\rightarrow x$ as $n\rightarrow \infty$, so $x=\sqrt{3\sqrt{5x}}$ $\Leftrightarrow$ $x^4=45x$ $\Leftrightarrow$ $x=3^{\frac{2}{3}}5^{\frac{1}{3}}$.

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I would have started with $$\sqrt{3\sqrt{5\sqrt{3\sqrt{5\cdots}}}}= \sqrt{\sqrt{3^2\cdot5\sqrt{\sqrt{3^2\cdot5\cdots}}}}$$ and then used the method that I have recently learned here in an answer to one the related questions.

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For completeness you might want to link to that answer, illustrating "the method" of how you would continue after the first step. –  TMM Jan 7 at 13:12
    
@TMM, Madrit Zhaku already posted an answer along those lines. As usual, the different kinds of answers there are repeated here ;) I only wanted to show the reduction. Btw, it has now occurred to my that one might, if one is very careful, want to show that $\sqrt{3\sqrt{\sqrt{5^2\cdot3\sqrt{\sqrt{5^2\cdot3\cdots}}}}}$ has the same limit. –  Carsten Schultz Jan 7 at 15:29

Let $x=\sqrt{3\sqrt{5\sqrt{3\sqrt{5\cdots}}}}$ or $x=\sqrt{3\sqrt{5x}}$.

Squaring both sides: $x^2=3\sqrt{5x}$, squaring again: $x^4=9\cdot5x$. $$x^4-45x = 0$$ $$x(x^3-45)=0$$

Since $x>0$, then $x^3=45$.

$x=\sqrt[3]{45}=3.55689...$

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You can use $\LaTeX$ to get nice-looking formulas. See an example of how I edited your answer (click 'edited X min ago' and select 'side-by-side markdown' there). –  Ruslan Jan 7 at 14:00

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