What is the sum from n=1 to infinity of (3/8)^{1-n} ?

The sum from n=1 to infinity of (3/8)^{1-n} is diverges

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sum from n=1 to infinity of (3/8)^{1-n}

n = 1 \sum \infty (\frac{3}{8})^{1 - n}

diverges

Solution steps

n = 1 \sum \infty (\frac{3}{8})^{1 - n}

Apply Series Ratio Test:

diverges

= diverges

x \to 1 lim (e^{6 x^{4} - 6 x})

f^{^{'}} (x) = 2^{x}

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

x \to 0 lim (ln (x) x)

\frac{\partial}{\partial x} (x^{2} y^{3} + 6 x yz - 2 yz)

What is the sum from n=1 to infinity of (3/8)^{1-n} ?
The sum from n=1 to infinity of (3/8)^{1-n} is diverges