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

The sum from n=1 to infinity of (-2+1)^n is diverges

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sum from n=1 to infinity of (-2+1)^n

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

diverges

Solution steps

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

Simplify

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

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

= n = 0 \sum \infty (- 1)^{n} - (- 1)^{0}

n = 0 \sum \infty (- 1)^{n} =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

n = 2 \sum \infty \frac{n ^{e}}{n ^{π}}

n = 1 \sum \infty - 4 (\frac{5}{8})^{n}

k = 1 \sum \infty \frac{9}{k} - \frac{9}{k + 1}

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

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

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