What is the sum from n=1 to infinity of 2(1/5)^{n-1} ?

The sum from n=1 to infinity of 2(1/5)^{n-1} is converges

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sum from n=1 to infinity of 2(1/5)^{n-1}

n = 1 \sum \infty 2 (\frac{1}{5})^{n - 1}

converges

Solution steps

n = 1 \sum \infty 2 (\frac{1}{5})^{n - 1}

Apply the constant multiplication rule:

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

= 2 \cdot n = 1 \sum \infty (\frac{1}{5})^{n - 1}

Apply Series Ratio Test:

converges

= 2 converges

= converges

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

y^{^{'}} + 3 x^{- 1} y = x^{2}

\int_{1}^{2} x^{4} ln (x) d x

\frac{\partial}{\partial x} (y^{2} tan (x y))

5 e^{2 x}

What is the sum from n=1 to infinity of 2(1/5)^{n-1} ?
The sum from n=1 to infinity of 2(1/5)^{n-1} is converges