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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Ratio Test:

converges

= 2 converges

= converges

n = 1 \sum \infty \frac{1}{8 n + 1}

n = 1 \sum \infty 4 ln (\frac{n}{n + 1})

n = 0 \sum \infty (\frac{1}{12})^{n}

n = 4 \sum \infty ln (\frac{n}{2})

n = 1 \sum \infty \frac{1}{9 + n ^{2}}

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