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

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

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

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

converges

Solution steps

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

n \to \infty lim (n (\frac{\frac{1}{2 ^{2^{n}}}}{\frac{1}{2 ^{2^{n + 1}}}} - 1)) = \infty

L > 1,

by the Raabe's test

= converges

y = π^{3 x + 1}

\frac{d z}{d z}

\frac{d}{d x} ((x + y + 1)^{2})

f (x) = cos (a^{9} + x^{9})

\int \frac{x}{25 - 4 x ^{2}} d x

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