What is the sum from n=1 to infinity of 5/(3^{-n)} ?

The sum from n=1 to infinity of 5/(3^{-n)} is diverges

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sum from n=1 to infinity of 5/(3^{-n)}

n = 1 \sum \infty \frac{5}{3 ^{- n}}

diverges

Solution steps

n = 1 \sum \infty \frac{5}{3 ^{- n}}

Apply the constant multiplication rule:

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

= 5 \cdot n = 1 \sum \infty \frac{1}{3 ^{- n}}

Apply Series Ratio Test:

diverges

= 5 diverges

= diverges

\frac{d}{d x} (\frac{4}{x ^{10}})

f (x) = e^{\frac{2}{x}}

y = arctan (\frac{4 x}{3})

\int \frac{1}{16 - x ^{2}} d x

{\frac{2 s - 1}{s ^{2} ( s + 1 ) ^{3}}}

What is the sum from n=1 to infinity of 5/(3^{-n)} ?
The sum from n=1 to infinity of 5/(3^{-n)} is diverges