What is the sum from n=0 to infinity of n*2^{n-1} ?

The sum from n=0 to infinity of n*2^{n-1} is diverges

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sum from n=0 to infinity of n*2^{n-1}

n = 0 \sum \infty n \cdot 2^{n - 1}

diverges

Solution steps

n = 0 \sum \infty n \cdot 2^{n - 1}

Simplify

2^{n - 1} : 2^{n} \cdot 2^{- 1}

= n = 0 \sum \infty n \cdot 2^{n} \cdot 2^{- 1}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 2^{- 1} \cdot n = 0 \sum \infty n \cdot 2^{n}

Apply exponent rule:

a^{- 1} = \frac{1}{a}

= \frac{1}{2} \cdot n = 0 \sum \infty n \cdot 2^{n}

Apply Series Ratio Test:

diverges

= \frac{1}{2} diverges

= diverges

What is the sum from n=0 to infinity of n*2^{n-1} ?
The sum from n=0 to infinity of n*2^{n-1} is diverges