What is the sum from n=0 to infinity of 4^{1/(2^n)} ?

The sum from n=0 to infinity of 4^{1/(2^n)} is diverges

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sum from n=0 to infinity of 4^{1/(2^n)}

n = 0 \sum \infty 4^{\frac{1}{2 ^{n}}}

diverges

Solution steps

n = 0 \sum \infty 4^{\frac{1}{2 ^{n}}}

Apply Series Divergence Test:

diverges

= diverges

y = \frac{1}{x}, y = \frac{1}{x ^{2}}, y = 4

x \to + \frac{π}{2} lim (sec^{2} (x))

3 x^{2} + \frac{16 x}{3} - 2

f (x) = \frac{x}{x ^{2} + 3}

\frac{( z - 2 )}{z - 1}

What is the sum from n=0 to infinity of 4^{1/(2^n)} ?
The sum from n=0 to infinity of 4^{1/(2^n)} is diverges