What is the sum from n=2 to infinity of 7/(10^{n-1)} ?

The sum from n=2 to infinity of 7/(10^{n-1)} is converges

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sum from n=2 to infinity of 7/(10^{n-1)}

n = 2 \sum \infty \frac{7}{1 0 ^{n - 1}}

converges

Solution steps

n = 2 \sum \infty \frac{7}{1 0 ^{n - 1}}

Apply the constant multiplication rule:

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

= 7 \cdot n = 2 \sum \infty \frac{1}{1 0 ^{n - 1}}

Apply Series Ratio Test:

converges

= 7 converges

= converges

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

\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

What is the sum from n=2 to infinity of 7/(10^{n-1)} ?
The sum from n=2 to infinity of 7/(10^{n-1)} is converges