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

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

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

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

n = 1 \sum \infty arccos (\frac{1}{n})

n = 2 \sum \infty \frac{n}{3 ^{n}}

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

n = 0 \sum \infty \frac{1}{n ^{e}}

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