What is the sum from n=1 to infinity of (3^n)/(e^n) ?

The sum from n=1 to infinity of (3^n)/(e^n) is diverges

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sum from n=1 to infinity of (3^n)/(e^n)

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

diverges

Solution steps

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

Apply Series Ratio Test:

diverges

= diverges

n = 0 \sum \infty 9 (\frac{8}{9})^{n}

n = 10 \sum \infty \frac{1}{n ln ( n )}

n = 1 \sum \infty 0. 1^{n}

i = 1 \sum \infty \frac{1}{i ( i + 1 )}

n = 0 \sum \infty \frac{6}{3 ^{n}}

What is the sum from n=1 to infinity of (3^n)/(e^n) ?
The sum from n=1 to infinity of (3^n)/(e^n) is diverges