What is the sum from n=1 to infinity of 0.75*0.5^{n-1} ?

The sum from n=1 to infinity of 0.75*0.5^{n-1} is converges

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sum from n=1 to infinity of 0.75*0.5^{n-1}

n = 1 \sum \infty 0.75 \cdot 0. 5^{n - 1}

converges

Solution steps

n = 1 \sum \infty 0.75 \cdot 0. 5^{n - 1}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 0.75 \cdot n = 1 \sum \infty 0. 5^{n - 1}

Apply Series Ratio Test:

converges

= 0.75 converges

= converges

\int e^{(2 x)} d x

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

\int sin^{3} (2 x) cos^{6} (2 x) d x

x = 4 y, x = 0, y = 4

\frac{e ^{x}}{2 x ^{2}}

What is the sum from n=1 to infinity of 0.75*0.5^{n-1} ?
The sum from n=1 to infinity of 0.75*0.5^{n-1} is converges