What is the sum from n=0 to infinity of 2(3)^n ?

The sum from n=0 to infinity of 2(3)^n is diverges

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sum from n=0 to infinity of 2(3)^n

n = 0 \sum \infty 2 (3)^{n}

diverges

Solution steps

n = 0 \sum \infty 2 \cdot 3^{n}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 2 \cdot n = 0 \sum \infty 3^{n}

Apply Series Geometric Test:

diverges

= 2 diverges

= diverges

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

n = 0 \sum \infty n^{3} (- 1)^{n}

n = 1 \sum \infty \frac{5}{e ^{n}}

n = 1 \sum \infty (sin (400))^{n}

n = 0 \sum \infty \frac{n !}{n + 1}

What is the sum from n=0 to infinity of 2(3)^n ?
The sum from n=0 to infinity of 2(3)^n is diverges