What is the sum from n=1 to infinity of 2^n3^{1-n} ?

The sum from n=1 to infinity of 2^n3^{1-n} is converges

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sum from n=1 to infinity of 2^n3^{1-n}

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

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

k = 1 \sum \infty (1 + \frac{2}{k})^{k}

n = 1 \sum \infty \frac{1}{n ^{0.8}}

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

n = 1 \sum \infty (\frac{6 π}{7})^{n}

What is the sum from n=1 to infinity of 2^n3^{1-n} ?
The sum from n=1 to infinity of 2^n3^{1-n} is converges