What is the sum from n=0 to infinity of n/(1-2n) ?

The sum from n=0 to infinity of n/(1-2n) is diverges

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sum from n=0 to infinity of n/(1-2n)

n = 0 \sum \infty \frac{n}{1 - 2 n}

diverges

Solution steps

n = 0 \sum \infty \frac{n}{1 - 2 n}

Factor

1 - 2 n : - (- 1 + 2 n)

= n = 0 \sum \infty \frac{n}{- ( - 1 + 2 n )}

Apply the constant multiplication rule:

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

= - n = 0 \sum \infty \frac{n}{- 1 + 2 n}

Apply Series Divergence Test:

diverges

= - diverges

= diverges

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

k = 3 \sum \infty 5 (\frac{1}{3})^{k}

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

n = 1 \sum \infty (\frac{5}{8})^{n - 1}

k = 2 \sum \infty (- 0.4)^{k}

What is the sum from n=0 to infinity of n/(1-2n) ?
The sum from n=0 to infinity of n/(1-2n) is diverges