What is the sum from n=1 to infinity of 1-3/(7^n) ?

The sum from n=1 to infinity of 1-3/(7^n) is diverges

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sum from n=1 to infinity of 1-3/(7^n)

n = 1 \sum \infty 1 - \frac{3}{7 ^{n}}

diverges

Solution steps

n = 1 \sum \infty 1 - \frac{3}{7 ^{n}}

Apply Series Divergence Test:

diverges

= diverges

\frac{d y}{d t} = \frac{1}{2} t

y^{^{''}} + 6 y^{^{'}} + 9 y = 12 e^{- 3 x}

\int 50 sin (120 π) x d x

x \cdot d y = d x, y (- 1) = 3

\frac{d}{d x} (7^{x} lo g_{7} (x))

What is the sum from n=1 to infinity of 1-3/(7^n) ?
The sum from n=1 to infinity of 1-3/(7^n) is diverges