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

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

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sum from n=1 to infinity of (-pi)^n

n = 1 \sum \infty (- π)^{n}

diverges

Solution steps

n = 1 \sum \infty (- π)^{n}

(- π)^{n} = 3.14159 \dots^{n} (- 1)^{n}

= n = 1 \sum \infty 3.14159 \dots^{n} (- 1)^{n}

Apply Series Ratio Test:

diverges

= diverges

n = 2 \sum \infty n e^{- 4 n}

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

n = 2 \sum \infty (\frac{3}{4})^{3 n}

n = 1 \sum \infty \frac{5 n}{6 n + 1}

n = 1 \sum \infty \frac{1}{6}

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