What is the sum from n=1 to infinity of 5/(4^{n+1)} ?

The sum from n=1 to infinity of 5/(4^{n+1)} is converges

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sum from n=1 to infinity of 5/(4^{n+1)}

n = 1 \sum \infty \frac{5}{4 ^{n + 1}}

converges

Solution steps

n = 1 \sum \infty \frac{5}{4 ^{n + 1}}

Apply the constant multiplication rule:

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

= 5 \cdot n = 1 \sum \infty \frac{1}{4 ^{n + 1}}

Apply Series Ratio Test:

converges

= 5 converges

= converges

\frac{d}{d x} (cos (6 x^{3}))

(1 + y^{2}) d x + (1 + x^{2}) d y = 0

t - 5

f (x) = \frac{8}{x ^{2}}, at x = 2

x \to 0 lim (\frac{2}{x + 5})

What is the sum from n=1 to infinity of 5/(4^{n+1)} ?
The sum from n=1 to infinity of 5/(4^{n+1)} is converges