What is the sum from n=0 to infinity of 4(1/3)^n ?

The sum from n=0 to infinity of 4(1/3)^n is 6

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sum from n=0 to infinity of 4(1/3)^n

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

6

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Geometric Test:

\frac{3}{2}

= 4 \cdot \frac{3}{2}

Simplify

4 \cdot \frac{3}{2} : 6

= 6

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

n = 2 \sum \infty \frac{7}{n ( n + 2 )}

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

n = 1 \sum \infty \frac{1}{( n + 1 ) n}

n = 1 \sum \infty sin (9)

What is the sum from n=0 to infinity of 4(1/3)^n ?
The sum from n=0 to infinity of 4(1/3)^n is 6