Question

prove :
√(2√(3√(4√(...))) < 3

Answer 1

all square roots?\[\sqrt{2\sqrt{3\sqrt{4\sqrt{...}}}}<3\]

Answer 2

yeah

Answer 3

I'm helpless on these things, but enjoy the show :)

Answer 4

:)

Answer 5

we must think of something like this

\[\sqrt{2\sqrt{3\sqrt{4\sqrt{...}}}}<\sqrt{2\sqrt{4\sqrt{8\sqrt{...}}}}=2^{\frac{1}{2}+\frac{1}{2}+\frac{3}{8}+... }=4\sqrt{2}\]

Answer 6

can you explain how\[2^{\frac{1}{2}+\frac{1}{2}+\frac{3}{8}+... }=4\sqrt{2}\]?

Answer 7

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

Answer 8

@KingGeorge
what do u think?

Answer 9

The only other possible method I know of would be to square both sides, solve for the square root, and find a pattern in the right side.

Answer 10

\[\sqrt{2\sqrt{3\sqrt{4\sqrt{...}}}}\overset{?}<3\]\[\sqrt{3\sqrt{4\sqrt{...}}}\overset{?}<\frac{9}{2}<3\]\[\vdots\]

Answer 11

I have my doubts about this way working however. @mukushla's method will probably be your best bet.

Answer 12

@TuringTest

************we must think of something like this

\[\sqrt{2\sqrt{3\sqrt{4\sqrt{...}}}}<\sqrt{2\sqrt{4\sqrt{8\sqrt{...}}}}=2^{\frac{1}{2}+\frac{1}{2}+\frac{3}{8}+... }=4\]

Answer 13

reverse induction will work here

Answer 14

yeah its working.......

let \[f(n)=\sqrt{m\sqrt{(m+1)\sqrt{(m+2)\sqrt{...\sqrt{n}}}}}\]
for \(m=n\) we have \(\sqrt{n}<n+1\) true...
let suppose this is true for \(m+1\) it means
\[\sqrt{(m+1)\sqrt{(m+2)\sqrt{(m+3)\sqrt{...\sqrt{n}}}}}<m+2\]
we show that its true for \(m\)
\[\sqrt{m\sqrt{(m+1)\sqrt{(m+2)\sqrt{...\sqrt{n}}}}}<\sqrt{m(m+2)}<m+1\]

so the statement is true for all \(n \in N\) let \(m=2\) we are done...

Answer 15

tnx @Neemo

the definition of \(f\) we assume that \(m\le n\)

and we want to show that for all \(m\le n\) this statement is true : \(f(n)<m+1\)

Answer 16

Oke, thanks mukushla and any one help me solved this problem...

Answer 17

yw :)