Hi Brilliant Community ,

Recently I was Solving Some Problems and i was perplexed by a expression

and I want to know if this expression is right .

\[\large \dots\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{2}}}}}\]
\[\large {= 2^{\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \dots}}\]
\[\large{= 2 ^{\frac{1}{∞}}}\]
\[\large{= 2^{0}}\]

\[\large{= 1}\]

This means that \(\dots\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{2}}}}}=1\)

\[\large{ 2 ^{\frac{1}{∞}}=1}\]
\[\large{2=1^{∞}}\]

Is this proof wrong ? and if I am wrong , tell me where it is

*(all help appreciated)*

Please **reshare** this note so that we can get better answers .

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## Comments

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TopNewestThe simple answer is yes, the proof is wrong. In fact, it is correct up until the very end. (or at least the underlying logic is)

The error is between the final two lines. \[2^{\frac{1}{\infty}} = 1 \\ (2^{\frac{1}{\infty}})^{\infty} = 1^{\infty}\] In this second line, both sides of the equation are indeterminate forms. That is \(1^{\infty} \neq 1.\) In fact it is mathematically incorrect really to algebraically put any number to the power of infinity; more properly, you would have to raise it to the power of a limit that approaches infinity. For this reason, to maintain accuracy, you must evaluate the inside exponent first; that is, \[(2^{\frac{1}{\infty}})^{\infty} = 1^{\infty}\] if a determinate form exists. (edit: to clarify, \((2^{\frac{1}{\infty}})^{\infty} \neq 2\))

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I want to correct you a bit

\(1^{\infty } = 1 \) when 1 on LHS is exact 1

Otherwise \(1^{\infty } \neq 1 \) when 1 is tending 1.

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You are thinking of the limit of \(1^n\) as \(n\) approaches infinity. In fact, the number \(1^{\infty}\) itself is indeterminate, therefore not \(1\) even if the base is exactly \(1.\) You could try to define it as the limit I described above, but you could define it as infinitely many other limits as well to get different values. If you don't believe me then check these:

Link 1 Link 2 Link 3

As you can see, wolfram alpha correctly describes it as indeterminate, and the other two links found different values for \(1^{\infty}.\) Remember, infinity is not a number; you can't just expect real numbers reasoning to hold at infinity.

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\(\displaystyle \lim_{x \to 0} (1+x)^{\frac{1}{x}}\)

The base in the above example is

tending1.I'll soon make a note about indeterminant form :)

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I guarantee you will not change my mind, but I do recommend that you research this as well; any reliable source will say that \(1^{\infty}\) is indeterminate.

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The problem with your proof is specially in the last two steps. Infinity is not a number and it can not be treated as a number.

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A very common mistake made is not understanding limits of sequences properly, and thinking that induction applies to "infinity".

As a starting example,

1. Prove that for all \( n \in \mathbb{N} \), \[ \bigcap_{i=1}^n ( 0 , \frac{1}{n} ) \neq \emptyset . \] 2. Prove that \[ \bigcap_{i=1}^ \infty ( 0 , \frac{1}{n} ) = \emptyset . \]

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