What is the sum from n=1 to infinity of 4/(3^n) ?

The sum from n=1 to infinity of 4/(3^n) is 2

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sum from n=1 to infinity of 4/(3^n)

n = 1 \sum \infty \frac{4}{3 ^{n}}

2

Solution steps

n = 1 \sum \infty \frac{4}{3 ^{n}}

Apply the constant multiplication rule:

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

= 4 \cdot n = 1 \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}

= 4 (n = 0 \sum \infty \frac{1}{3 ^{n}} - \frac{1}{3 ^{0}})

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

= 4 (\frac{3}{2} - \frac{1}{3 ^{0}})

Simplify

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

= 2

\int_{6}^{8} \frac{8}{( x - 6 ) ^{3}} d x

x \to - 7 lim (2 x^{2} - 6 x)

(x + 9) (7 x + 2)

f (x) = 3 x, (3, 3)

\frac{2 s + 2}{s ^{2} + 4 s + 5}

What is the sum from n=1 to infinity of 4/(3^n) ?
The sum from n=1 to infinity of 4/(3^n) is 2