What is the sum from n=1 to infinity of 7/(4^{n-1)} ?

The sum from n=1 to infinity of 7/(4^{n-1)} is converges

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sum from n=1 to infinity of 7/(4^{n-1)}

n = 1 \sum \infty \frac{7}{4 ^{n - 1}}

converges

Solution steps

n = 1 \sum \infty \frac{7}{4 ^{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}{4 ^{n - 1}}

Apply Series Ratio Test:

converges

= 7 converges

= converges

\int x - 2 d x

\frac{d y}{d x} = y (4 x^{2} + 6)

y^{^{''}} - 6 y^{^{'}} + 3 y = 0, y (0) = 0, y^{^{'}} (0) = 5

y^{^{'}} = \frac{( 3 x ^{2} - e ^{x} )}{( 2 y - 5 )}

h \to 0 lim (\frac{( 3 + h ) ^{2} - 9}{h})

What is the sum from n=1 to infinity of 7/(4^{n-1)} ?
The sum from n=1 to infinity of 7/(4^{n-1)} is converges