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

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

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

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

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 = 1 \sum \infty 4^{n - 1}

n = 1 \sum \infty 2 (0.1)^{n - 1}

n = 1 \sum \infty \frac{3}{( 2 n + 5 ) ^{3}}

n = 1 \sum \infty \frac{3 n}{n ^{2} - 7}

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

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