What is the sum from n=2 to infinity of 2/(3^n) ?

The sum from n=2 to infinity of 2/(3^n) is 1/3

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sum from n=2 to infinity of 2/(3^n)

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

\frac{1}{3}

Solution steps

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

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 2 \cdot n = 2 \sum \infty \frac{1}{3 ^{n}}

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

= 2 (n = 0 \sum \infty \frac{1}{3 ^{n}} - \frac{1}{3 ^{0}} - \frac{1}{3 ^{1}})

n = 0 \sum \infty \frac{1}{3 ^{n}} = \frac{3}{2}

= 2 (\frac{3}{2} - \frac{1}{3 ^{0}} - \frac{1}{3 ^{1}})

Simplify

2 (\frac{3}{2} - \frac{1}{3 ^{0}} - \frac{1}{3 ^{1}}) : \frac{1}{3}

= \frac{1}{3}

\int \frac{2 x}{( x ^{2} + 2 ) ln ( x ^{2} + 2 )} d x

\frac{d y}{d t} - y = 1

\frac{d}{d x} (e^{2 x} ln (3 x))

\frac{d}{d x} (x^{3} + 4 x^{2} + 2)

\frac{d}{d x} (x^{5} - 4 x^{3})

What is the sum from n=2 to infinity of 2/(3^n) ?
The sum from n=2 to infinity of 2/(3^n) is 1/3