What is the sum from n=1 to infinity of (4^n)/(2^n) ?

The sum from n=1 to infinity of (4^n)/(2^n) is diverges

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sum from n=1 to infinity of (4^n)/(2^n)

n = 1 \sum \infty \frac{4 ^{n}}{2 ^{n}}

diverges

Solution steps

n = 1 \sum \infty \frac{4 ^{n}}{2 ^{n}}

Simplify

\frac{4 ^{n}}{2 ^{n}} : 2^{n}

= n = 1 \sum \infty 2^{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 2^{n} - 2^{0}

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

x y - 3 y^{2} = 8, (14, 4)

\int x e^{- x} d x

\frac{d}{d x} (\frac{2 x}{ln ( x )})

\frac{d}{d x} (\frac{x ^{3}}{1 - x ^{3}})

\frac{\partial}{\partial x} (x + e^{z} + y)

What is the sum from n=1 to infinity of (4^n)/(2^n) ?
The sum from n=1 to infinity of (4^n)/(2^n) is diverges