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

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

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

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

diverges

Solution steps

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

Factor

1 - n : - (- 1 + n)

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

Apply the constant multiplication rule:

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

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

Apply Series Comparison Test:

diverges

= - diverges

= diverges

n = 0 \sum \infty 6^{- n}

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

n = 1 \sum \infty (\frac{8}{7})^{n}

n = 1 \sum \infty \frac{2 n + 1}{n + 3}

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

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