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

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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

y^{^{'}} = 3 x y + 5 y

sin (3 x) d x + 2 y cos^{3} (3 x) d y = 0

\frac{d y}{d x} = \frac{3 x ^{2} + 4 x + 2}{2 ( y - 1 )}

\frac{d v}{d t} = 1 - 0.4 v

\frac{3 x ^{2} + 1}{2 y - 3} = \frac{d y}{d x}

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