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

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

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

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

diverges

Solution steps

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

n = 0 \sum \infty (- 2)^{n} =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\frac{( s + 2 )}{s ^{2} + 7 s + 12}

\frac{\partial}{\partial x} (sin (3 x + 2 y + z))

y^{^{''}} - 2 \cdot y = 981, y^{^{'}} (0) = 0, y (0) = 0

\frac{\partial}{\partial x} (\frac{( 3 x + y )}{z})

\frac{\partial}{\partial x} (sin (\frac{x - y}{x + y}))

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