What is the sum from n=0 to infinity of n2^n ?

The sum from n=0 to infinity of n2^n is diverges

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sum from n=0 to infinity of n2^n

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

diverges

Solution steps

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

Apply Series Ratio Test:

diverges

= diverges

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

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

n = 1 \sum \infty \frac{1}{n ^{5} + 8}

n = 0 \sum \infty cos (\frac{n}{2})

n = 0 \sum \infty (\frac{- 1}{3})^{n}

What is the sum from n=0 to infinity of n2^n ?
The sum from n=0 to infinity of n2^n is diverges