What is the sum from n=1 to infinity of (2n+1)/(6^n) ?

The sum from n=1 to infinity of (2n+1)/(6^n) is converges

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sum from n=1 to infinity of (2n+1)/(6^n)

n = 1 \sum \infty \frac{2 n + 1}{6 ^{n}}

converges

Solution steps

n = 1 \sum \infty \frac{2 n + 1}{6 ^{n}}

Apply Series Ratio Test:

converges

= converges

n = 0 \sum \infty \frac{4}{3 - n}

n = 0 \sum \infty \frac{1}{n ^{4} + n ^{2}}

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

k = 2 \sum \infty (- 0.35)^{k}

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

What is the sum from n=1 to infinity of (2n+1)/(6^n) ?
The sum from n=1 to infinity of (2n+1)/(6^n) is converges