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

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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

n = 1 \sum \infty ln (1 + n)

n = 0 \sum \infty (\frac{5 π}{e})^{n}

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

n = 5 \sum \infty \frac{1}{n !}

n = 0 \sum \infty \frac{3}{3 - 6 n}

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