What is the sum from n=1 to infinity of 1-7/(4^n) ?

The sum from n=1 to infinity of 1-7/(4^n) is diverges

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sum from n=1 to infinity of 1-7/(4^n)

n = 1 \sum \infty 1 - \frac{7}{4 ^{n}}

diverges

Solution steps

n = 1 \sum \infty 1 - \frac{7}{4 ^{n}}

Apply Series Divergence Test:

diverges

= diverges

\frac{\partial}{\partial x} (8 - x^{2} - y^{2})

24 y^{^{''}} - 2 y^{^{'}} - 15 y = 0

\frac{d}{d t} (cos^{2} (e^{c o s^{2} (t)}))

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

\frac{d y}{d x} = \frac{cos ( 16 x )}{e ^{16 y}}

What is the sum from n=1 to infinity of 1-7/(4^n) ?
The sum from n=1 to infinity of 1-7/(4^n) is diverges