What is the sum from n=1 to infinity of 7/(2^{n-1)} ?

The sum from n=1 to infinity of 7/(2^{n-1)} is converges

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sum from n=1 to infinity of 7/(2^{n-1)}

n = 1 \sum \infty \frac{7}{2 ^{n - 1}}

converges

Solution steps

n = 1 \sum \infty \frac{7}{2 ^{n - 1}}

Apply the constant multiplication rule:

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

= 7 \cdot n = 1 \sum \infty \frac{1}{2 ^{n - 1}}

Apply Series Ratio Test:

converges

= 7 converges

= converges

\int \frac{16}{sec ( 2 x ) - 1} d x

\frac{d}{d x} (2 e^{2 x} + 3 e^{3 x})

x \to a lim ([f (x)]^{g (x)})

t \to 0 lim (\frac{3}{t} - \frac{3}{e ^{t} - 1})

y^{^{'}} + y = t e^{- t} + 1

What is the sum from n=1 to infinity of 7/(2^{n-1)} ?
The sum from n=1 to infinity of 7/(2^{n-1)} is converges