What is the sum from n=0 to infinity of-2^n ?

The sum from n=0 to infinity of-2^n is diverges

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sum from n=0 to infinity of-2^n

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Geometric Test:

diverges

= - diverges

= diverges

\frac{d}{d x} (x^{3} (x + 2))

\int 5 (5 x - 1)^{3} d x

\int e^{x} + x cos (y) d y

n = 1 \sum \infty n^{3} \cdot e^{- n^{2}}

x \to 4 lim (5 x^{2} - 2 x)

What is the sum from n=0 to infinity of-2^n ?
The sum from n=0 to infinity of-2^n is diverges