Can we calculate i\sqrt { i\sqrt { i\sqrt { \cdots } } } i\sqrt { i\sqrt { i\sqrt { \cdots } } }?

It might be obvious that 2\sqrt { 2\sqrt { 2\sqrt { 2\sqrt { 2\sqrt { 2\sqrt { \cdots } } } } } } equals 4. So what about i\sqrt { i\sqrt { i\sqrt { i\sqrt { i\sqrt { i\sqrt { \cdots } } } } } } \text{ ?} The answer might be -1, but I’m not sure as i is not a real number. Can anyone help?

Answer

\begin{eqnarray*}
x= a\sqrt { a\sqrt { a\sqrt { a\sqrt { a\sqrt { a\sqrt { \cdots } } } } } } \\
x=a^{ 1+1/2+1/4+1/8+\cdots} \\
x=a^2
\end{eqnarray*}

So it would seem that
\begin{eqnarray*}
i\sqrt { i\sqrt { i\sqrt { i\sqrt { i\sqrt { i\sqrt { \cdots } } } } } }=\color{red}{-1}.
\end{eqnarray*}

Attribution
Source : Link , Question Author : Community , Answer Author : Donald Splutterwit

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