What is the sum from n=0 to infinity of (n2^n)/(3^n) ?

The sum from n=0 to infinity of (n2^n)/(3^n) is converges

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sum from n=0 to infinity of (n2^n)/(3^n)

n = 0 \sum \infty \frac{n 2 ^{n}}{3 ^{n}}

converges

Solution steps

n = 0 \sum \infty \frac{n \cdot 2 ^{n}}{3 ^{n}}

Apply Series Ratio Test:

converges

= converges

y = x^{3} (2 x - 1)^{5}

\frac{\partial}{\partial x} (\frac{x + y}{1 - x y})

f (x) = x^{3} - 3 x

(x^{2} - 5) (x + 7)

\int sin (11 x) d x

What is the sum from n=0 to infinity of (n2^n)/(3^n) ?
The sum from n=0 to infinity of (n2^n)/(3^n) is converges