What is the sum from n=1 to infinity of 2(1/4)^{n-1} ?

The sum from n=1 to infinity of 2(1/4)^{n-1} is converges

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sum from n=1 to infinity of 2(1/4)^{n-1}

n = 1 \sum \infty 2 (\frac{1}{4})^{n - 1}

converges

Solution steps

n = 1 \sum \infty 2 (\frac{1}{4})^{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}{4})^{n - 1}

Apply Series Ratio Test:

converges

= 2 converges

= converges

n = 1 \sum \infty \frac{4}{n ^{- 2.3}}

n = 0 \sum \infty (- 0.2)^{n}

n = 0 \sum \infty \frac{1}{( 1 + n )}

n = 0 \sum \infty cos (\frac{3}{n ^{2}})

n = 2 \sum \infty \frac{1}{( n ln ( n ) )}

What is the sum from n=1 to infinity of 2(1/4)^{n-1} ?
The sum from n=1 to infinity of 2(1/4)^{n-1} is converges