What is the sum from n=0 to infinity of-3^n ?

The sum from n=0 to infinity of-3^n is diverges

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sum from n=0 to infinity of-3^n

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

diverges

Solution steps

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

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

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

Apply Series Geometric Test:

diverges

= - diverges

= diverges

\frac{7}{x}

\int \frac{1}{- x ^{2} + 4 x - 1} d x

y = x^{2} - 12 x + 4

\int e^{t} sin (3 t) d t

2 tan^{2} (4 x)

What is the sum from n=0 to infinity of-3^n ?
The sum from n=0 to infinity of-3^n is diverges