What is the sum from n=1 to infinity of (-3)^{n-1} ?

The sum from n=1 to infinity of (-3)^{n-1} is diverges

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sum from n=1 to infinity of (-3)^{n-1}

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

diverges

Solution steps

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

Apply Series Ratio Test:

diverges

= diverges

n = 4 \sum \infty \frac{1}{n ^{2} - 4 n + 3}

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

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

n = 1 \sum \infty \frac{5}{( ( - 4 ) ^{n} )}

k = 0 \sum \infty (\frac{1}{10})^{k}

What is the sum from n=1 to infinity of (-3)^{n-1} ?
The sum from n=1 to infinity of (-3)^{n-1} is diverges