What is the sum from n=1 to infinity of 6(0.4)^{n-1} ?

The sum from n=1 to infinity of 6(0.4)^{n-1} is converges

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sum from n=1 to infinity of 6(0.4)^{n-1}

n = 1 \sum \infty 6 (0.4)^{n - 1}

converges

Solution steps

n = 1 \sum \infty 6 \cdot 0. 4^{n - 1}

Apply the constant multiplication rule:

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

= 6 \cdot n = 1 \sum \infty 0. 4^{n - 1}

Apply Series Ratio Test:

converges

= 6 converges

= converges

n = 2 \sum \infty \frac{9}{n ln ( n )}

n = 0 \sum \infty (- \frac{7}{9})^{n}

n = 0 \sum \infty ln (1 + n^{2})

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

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

What is the sum from n=1 to infinity of 6(0.4)^{n-1} ?
The sum from n=1 to infinity of 6(0.4)^{n-1} is converges