What is the sum from n=1 to infinity of (n2^n)/(3^n) ?

The sum from n=1 to infinity of (n2^n)/(3^n) is converges

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sum from n=1 to infinity of (n2^n)/(3^n)

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

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

n = 1 \sum \infty \frac{n + 5}{n ^{2}}

n = 1 \sum \infty \frac{7 ^{n}}{n}

n = 1 \sum \infty [1 + (- 1)^{n}]

n = 1 \sum \infty \frac{4}{n ln ( n )}

What is the sum from n=1 to infinity of (n2^n)/(3^n) ?
The sum from n=1 to infinity of (n2^n)/(3^n) is converges