What is the sum from n=1 to infinity of (15/4)^{n+1} ?

The sum from n=1 to infinity of (15/4)^{n+1} is diverges

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sum from n=1 to infinity of (15/4)^{n+1}

n = 1 \sum \infty (\frac{15}{4})^{n + 1}

diverges

Solution steps

n = 1 \sum \infty (\frac{15}{4})^{n + 1}

Apply Series Ratio Test:

diverges

= diverges

\frac{\partial}{\partial x} (x^{y^{3}})

9 x - \frac{7}{x}

x \to - 9 lim (\frac{1}{( x + 9 ) ^{2}})

\int cos^{3} (3 x) sin^{- 2} (3 x) d x

x \to - \infty lim (19)

What is the sum from n=1 to infinity of (15/4)^{n+1} ?
The sum from n=1 to infinity of (15/4)^{n+1} is diverges