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

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

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

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

y = (2^{x})^{8}

\int (x^{2} + 2)^{4} 2 x d x

4 y^{^{''}} + 4 y^{^{'}} + 8 y = 0

x \to 0 lim (cos (9 x))

n = 1 \sum \infty n 2^{- n}

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