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

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

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

n = 0 \sum \infty 2^{n} \cdot (\frac{1}{4})^{l n (2^{n})}

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

\frac{d ^{2} y}{d x ^{2}} = 4 e^{- x}, y (0) = 3, y^{^{'}} (0) = 2

\frac{1}{2 r} + \frac{1}{3 r ^{\frac{2}{3}}}

y \to \infty lim (\frac{1}{5} (2 y^{2} + 3 y) x)

2 - x^{2}, 10 - 6 x

\int_{0}^{0.4} 300 x d x

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