What is the sum from n=1 to infinity of (4^n)/(n3^n) ?

The sum from n=1 to infinity of (4^n)/(n3^n) is diverges

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sum from n=1 to infinity of (4^n)/(n3^n)

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

diverges

Solution steps

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

Apply Series Ratio Test:

diverges

= diverges

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

n = 0 \sum \infty \frac{1}{3 e ^{n}}

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

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

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

What is the sum from n=1 to infinity of (4^n)/(n3^n) ?
The sum from n=1 to infinity of (4^n)/(n3^n) is diverges