What is the sum from n=0 to infinity of 5/(4^{n+1)} ?

The sum from n=0 to infinity of 5/(4^{n+1)} is converges

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sum from n=0 to infinity of 5/(4^{n+1)}

n = 0 \sum \infty \frac{5}{4 ^{n + 1}}

converges

Solution steps

n = 0 \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 = 0 \sum \infty \frac{1}{4 ^{n + 1}}

Apply Series Ratio Test:

converges

= 5 converges

= converges

x \to o lim (\frac{sin ( 4 x )}{x})

f^{^{'}} (x) = \frac{1}{x}

f (x) = 1 + 3 x

y^{^{''}} - 9 y = e^{- t} + 1

\int (sec^{2} (x) - tan^{2} (x)) d x

What is the sum from n=0 to infinity of 5/(4^{n+1)} ?
The sum from n=0 to infinity of 5/(4^{n+1)} is converges