What is the sum from n=1 to infinity of (8/5)^{n-1} ?

The sum from n=1 to infinity of (8/5)^{n-1} is diverges

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sum from n=1 to infinity of (8/5)^{n-1}

n = 1 \sum \infty (\frac{8}{5})^{n - 1}

diverges

Solution steps

n = 1 \sum \infty (\frac{8}{5})^{n - 1}

Apply Series Ratio Test:

diverges

= diverges

(t^{2} + 1) f^{^{'}} (t) + [f (t)]^{2} + 1 = 0, f (4) = e

\int \frac{x ^{3}}{x} d x

\int_{0}^{2} 5 x^{3} x^{2} + 4 d x

x \to 3 - lim (c x^{2} + 4 x)

\int \frac{x ^{2} + 5 x + 1}{( x ^{2} + 1 ) ^{2}} d x

What is the sum from n=1 to infinity of (8/5)^{n-1} ?
The sum from n=1 to infinity of (8/5)^{n-1} is diverges