What is the sum from n=0 to infinity of 3(2/6)^{n+2} ?

The sum from n=0 to infinity of 3(2/6)^{n+2} is converges

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sum from n=0 to infinity of 3(2/6)^{n+2}

n = 0 \sum \infty 3 (\frac{2}{6})^{n + 2}

converges

Solution steps

n = 0 \sum \infty 3 (\frac{2}{6})^{n + 2}

Apply the constant multiplication rule:

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

= 3 \cdot n = 0 \sum \infty (\frac{2}{6})^{n + 2}

Apply Series Ratio Test:

converges

= 3 converges

= converges

k = 1 \sum \infty \frac{k !}{10 4 ^{k}}

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

n = 2 \sum \infty (\frac{3}{7})^{4 n}

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

n = 1 \sum \infty \frac{6 n}{4 n + 1}

What is the sum from n=0 to infinity of 3(2/6)^{n+2} ?
The sum from n=0 to infinity of 3(2/6)^{n+2} is converges