What is the sum from n=1 to infinity of 2(0.4)^{n-1} ?

The sum from n=1 to infinity of 2(0.4)^{n-1} is converges

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sum from n=1 to infinity of 2(0.4)^{n-1}

n = 1 \sum \infty 2 (0.4)^{n - 1}

converges

Solution steps

n = 1 \sum \infty 2 \cdot 0. 4^{n - 1}

Apply the constant multiplication rule:

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

= 2 \cdot n = 1 \sum \infty 0. 4^{n - 1}

Apply Series Ratio Test:

converges

= 2 converges

= converges

\int x (x^{2} - 36)^{\frac{1}{2}} d x

\frac{1}{( 3 - x ) ^{2}} + \frac{1}{x}, 2

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

y = x^{2} + 4, (- 2, 8)

f (x) = 3 x^{\frac{2}{3}} - 2 x

What is the sum from n=1 to infinity of 2(0.4)^{n-1} ?
The sum from n=1 to infinity of 2(0.4)^{n-1} is converges