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

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

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

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

diverges

Solution steps

n = 1 \sum \infty (- 3)^{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 (- 3)^{n} - (- 3)^{0}

n = 0 \sum \infty (- 3)^{n} =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\frac{d y}{d x} = \frac{x - 3}{y + 5}

n = 1 \sum \infty 3 \cdot (\frac{1}{5})^{n}

n \to \infty lim (n^{2} e^{- 3 n})

\frac{( s - 3 )}{( s ^{2} + 4 s + 13 )}

y^{^{'}} - 7 y = - x

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