What is the sum from n=1 to infinity of 5/(6^{n-1)} ?

The sum from n=1 to infinity of 5/(6^{n-1)} is converges

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sum from n=1 to infinity of 5/(6^{n-1)}

n = 1 \sum \infty \frac{5}{6 ^{n - 1}}

converges

Solution steps

n = 1 \sum \infty \frac{5}{6 ^{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}{6 ^{n - 1}}

Apply Series Ratio Test:

converges

= 5 converges

= converges

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

n = 2 \sum \infty \frac{1}{( n ^{2} - 1 )}

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

n = 0 \sum \infty \frac{3 ^{n}}{2}

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

What is the sum from n=1 to infinity of 5/(6^{n-1)} ?
The sum from n=1 to infinity of 5/(6^{n-1)} is converges