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Let $n$ be the integer. Prove that $$\sqrt{2\sqrt{4\sqrt{8....\sqrt{2^n}}}} \leqslant n+1$$

SOURCE: BANGLADESH MATH OLYMPIAD

I am a new beginner at the infinite radical and sequence. I don't know about its basic conception.

My Attempt:

After seeing that problem, out of curiosity, I thought that I should calculate the multiplication of left side term.

Than I got something like this

$2^{\frac{1}{2}}$ × $2^{\frac{1}{2}}$ × $2^{\frac{3}{8}}$... × $2^{\frac{k}{2n}} = 2^{\left(\frac{1}{2}+\frac{1}{2}+\frac{3}{8}+...+\frac{k}{2n}\right)}$, where I denoted $2n$ = $2^\text{k}$.

But here I got stuck. The above sequence doesn't follow the pattern. Than how would I get the summation of the sequence?

Moreover, how should I approach to prove the above condition for the integer value of $n$? Any kind of help or clue will be greatly appreciated by this novice and new beginner. Thanks in advance.

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  • $\begingroup$ The question as it is written is unreadable. What does $\sqrt{4} \sqrt{8} \cdots \sqrt{2}$ mean? Is this supposed to be $\sqrt{4} \sqrt{8} \cdots \sqrt{2^n}$? $\endgroup$
    – parsiad
    Commented Feb 17, 2019 at 5:12
  • $\begingroup$ I edited to reflect what you likely meant instead of what you actually typed. I suspect your radicals are actually nested. If this is not the case, please edit to reflect what you actually meant. $\endgroup$
    – Cameron Williams
    Commented Feb 17, 2019 at 5:14
  • $\begingroup$ @CameronWilliams No my question is correct. It was as same described as in the source. I have again checked it out. But was this an absurd and illogical problem that can't be solved? The question maybe wrong but I was unable to trace out the mistake. But no problem! Your suggested edition is appropriate for learning. $\endgroup$
    – Anirban Niloy
    Commented Feb 17, 2019 at 5:50

1 Answer 1

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It's $$2^{\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+...+\frac{n}{2^n}}=2^{x(x+x^2+...+x^{n})'}_{x=\frac{1}{2}}=2^{x\left(\frac{x(x^n-1)}{x-1}\right)'}_{x=\frac{1}{2}}=2^{x\left(\frac{x^{n+1}-x}{x-1}\right)'}_{x=\frac{1}{2}}=$$ $$=2^{x\cdot\frac{((n+1)x^n-1)(x-1)-x^{n+1}+x}{(x-1)^2}}_{x=\frac{1}{2}}=2^{x\cdot\frac{nx^{n+1}-(n+1)x^n+1}{(x-1)^2}}_{x=\frac{1}{2}}=2^{2\left(n\left(\frac{1}{2}\right)^{n+1}-(n+1)\left(\frac{1}{2}\right)^n+1\right)}=$$ $$=2^{\frac{n}{2^n}-\frac{n+1}{2^{n-1}}+2}=2^{2-\frac{n+2}{2^n}}.$$ Thus, it's enough to prove that $$2^{2-\frac{n+2}{2^n}}\leq n+1,$$ which is obviously true for $n\geq3.$

But easy to check that for $n=1$ and for $n=2$ our inequality is also true.

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  • $\begingroup$ I got that. From there, I can approach. But at first, let me try with $n=1$ and $n=2$. 😅😅😅 $\endgroup$
    – Anirban Niloy
    Commented Feb 17, 2019 at 6:56

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