What is the sum from n=0 to infinity of 2^n3^{n-2} ?

The sum from n=0 to infinity of 2^n3^{n-2} is diverges

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sum from n=0 to infinity of 2^n3^{n-2}

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

diverges

Solution steps

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

Simplify

3^{n - 2} : 3^{n} \cdot 3^{- 2}

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

Apply the constant multiplication rule:

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

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

3^{- 2} = \frac{1}{9}

= \frac{1}{9} \cdot n = 0 \sum \infty 2^{n} \cdot 3^{n}

Simplify

2^{n} \cdot 3^{n} : 6^{n}

= \frac{1}{9} \cdot n = 0 \sum \infty 6^{n}

Apply Series Geometric Test:

diverges

= \frac{1}{9} diverges

= diverges

n = 5 \sum \infty \frac{ln ( n )}{n}

n = 1 \sum \infty 3^{- 3 n}

n = 3 \sum \infty 4 (\frac{1}{2})^{n + 2}

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

n = 1 \sum \infty 5 n^{2} e^{- n^{3}}

What is the sum from n=0 to infinity of 2^n3^{n-2} ?
The sum from n=0 to infinity of 2^n3^{n-2} is diverges