What is the sum from n=1 to infinity of (2^n3^n)/n ?

The sum from n=1 to infinity of (2^n3^n)/n is diverges

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sum from n=1 to infinity of (2^n3^n)/n

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

diverges

Solution steps

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

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

= n = 1 \sum \infty \frac{6 ^{n}}{n}

Apply Series Ratio Test:

diverges

= diverges

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

n = 1 \sum \infty \frac{π}{n ^{2}}

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

i = 1 \sum \infty 4 (\frac{2}{5})^{i - 1}

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

What is the sum from n=1 to infinity of (2^n3^n)/n ?
The sum from n=1 to infinity of (2^n3^n)/n is diverges