What is the sum from n=1 to infinity of (-4^n)/(4^n) ?

The sum from n=1 to infinity of (-4^n)/(4^n) is diverges

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sum from n=1 to infinity of (-4^n)/(4^n)

n = 1 \sum \infty \frac{- 4 ^{n}}{4 ^{n}}

diverges

Solution steps

n = 1 \sum \infty \frac{- 4 ^{n}}{4 ^{n}}

Apply the constant multiplication rule:

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

= - n = 1 \sum \infty \frac{4 ^{n}}{4 ^{n}}

Simplify

= - n = 1 \sum \infty 1

Every infinite sum of a non-zero constant diverges

= - diverges

= diverges

\int_{0}^{\infty} e^{- 0.8 x} d x

y^{^{''}} + 8 y^{^{'}} + 32 y = 32, y (0) = 2, y^{^{'}} (0) = - 20

\int cos (x) cos (2 x) cos (3 x) d x

\int_{0}^{\frac{3}{2} 3} \frac{x ^{3}}{9 - x ^{2}} d x

y = \frac{e ^{x} - e ^{- x}}{2}

What is the sum from n=1 to infinity of (-4^n)/(4^n) ?
The sum from n=1 to infinity of (-4^n)/(4^n) is diverges