What is the sum from n=1 to infinity of (n+1)/(3^n) ?

The sum from n=1 to infinity of (n+1)/(3^n) is converges

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sum from n=1 to infinity of (n+1)/(3^n)

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

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

n = 1 \sum \infty \frac{6 ^{n}}{n !}

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

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

n = 9 \sum \infty \frac{n !}{n ^{n}}

What is the sum from n=1 to infinity of (n+1)/(3^n) ?
The sum from n=1 to infinity of (n+1)/(3^n) is converges