What is the sum from n=1 to infinity of (-0.75)^n ?

The sum from n=1 to infinity of (-0.75)^n is converges

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sum from n=1 to infinity of (-0.75)^n

n = 1 \sum \infty (- 0.75)^{n}

converges

Solution steps

n = 1 \sum \infty (- 0.75)^{n}

Apply Series Ratio Test:

converges

= converges

\int x (x - 5) d x

y = 2 x^{2} ln (x), y = 8 ln (x)

\int \frac{48}{( t + 1 ) ^{4}} - 3 d t

\frac{d y}{d x} = \frac{x ^{7} - 8 y}{x}

\int_{0}^{1} e^{2 x} d x

What is the sum from n=1 to infinity of (-0.75)^n ?
The sum from n=1 to infinity of (-0.75)^n is converges