What is the sum from n=0 to infinity of 2^n2^n ?

The sum from n=0 to infinity of 2^n2^n is diverges

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sum from n=0 to infinity of 2^n2^n

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

diverges

Solution steps

n = 0 \sum \infty 2^{n} \cdot 2^{n}

Simplify

2^{n} \cdot 2^{n} : 4^{n}

= n = 0 \sum \infty 4^{n}

Apply Series Geometric Test:

diverges

= diverges

\frac{\partial}{\partial z} (x^{2} y^{2} z^{2})

x^{2} - 3 x + 6, at x = 1

\frac{d}{d x} (7 x^{2} - \frac{3}{x} + 2 x)

\frac{d}{d x} (\frac{x - 3}{x + 3})

x^{2} + 7

What is the sum from n=0 to infinity of 2^n2^n ?
The sum from n=0 to infinity of 2^n2^n is diverges