What is the sum from n=0 to infinity of-1/(3^n) ?

The sum from n=0 to infinity of-1/(3^n) is -3/2

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sum from n=0 to infinity of-1/(3^n)

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

- \frac{3}{2}

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Geometric Test:

\frac{3}{2}

= - \frac{3}{2}

x \to 0 lim (\frac{1}{x ^{2} + 1})

\frac{d}{d x} (sec^{2} (x) - 4)

\frac{d y}{d x} = \frac{1}{x ^{2}}

arcsin (\frac{x}{5})

x \to - 1 - lim (\frac{6}{x + 1})

What is the sum from n=0 to infinity of-1/(3^n) ?
The sum from n=0 to infinity of-1/(3^n) is -3/2