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

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

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

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

k = 2 \sum \infty (\frac{4}{5})^{3 k}

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

n = 1 \sum \infty 12 (- \frac{5}{8})^{n}

n = 2 \sum \infty 7 (- \frac{3}{8})^{n}

n = 2 \sum \infty (\frac{3}{5})^{4 n}

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