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

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

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

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

diverges

Solution steps

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

Factor

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

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

Apply the constant multiplication rule:

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

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

Apply Series Integral Test:

diverges

= - diverges

= diverges

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

n = 1 \sum \infty 4 (\frac{1}{3})^{n - 1}

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

n = 1 \sum \infty [53 (\frac{1}{2})^{n}]

n = 0 \sum \infty \frac{8}{n ( n + 2 )}

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