What is the sum from n=2 to infinity of (-0.75)^n ?

The sum from n=2 to infinity of (-0.75)^n is converges

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sum from n=2 to infinity of (-0.75)^n

n = 2 \sum \infty (- 0.75)^{n}

converges

Solution steps

n = 2 \sum \infty (- 0.75)^{n}

Apply Series Ratio Test:

converges

= converges

\frac{1}{x ^{2}} 0

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

\int \frac{e ^{5 x} + 5 e ^{x} + 7}{e ^{x}} d x

f (x) = 5 cot (x), at x = \frac{π}{4}

\frac{d y}{d x}, x + y = x^{2}

What is the sum from n=2 to infinity of (-0.75)^n ?
The sum from n=2 to infinity of (-0.75)^n is converges