What is the sum from n=0 to infinity of 2^n1^n ?

The sum from n=0 to infinity of 2^n1^n is diverges

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sum from n=0 to infinity of 2^n1^n

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

diverges

Solution steps

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

Simplify

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

Apply Series Geometric Test:

diverges

= diverges

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

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

\frac{d}{d x} (\frac{1}{( x ^{2} - 1 )})

- 45 y^{^{''}} + 35 y^{^{'}} = 0, y (2) = 2, y^{^{'}} (- 1) = - 1

(x + e^{y}) d y - d x = 0

What is the sum from n=0 to infinity of 2^n1^n ?
The sum from n=0 to infinity of 2^n1^n is diverges