What is the sum from n=1 to infinity of 5(1/3)^{n-1} ?

The sum from n=1 to infinity of 5(1/3)^{n-1} is converges

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sum from n=1 to infinity of 5(1/3)^{n-1}

n = 1 \sum \infty 5 (\frac{1}{3})^{n - 1}

converges

Solution steps

n = 1 \sum \infty 5 (\frac{1}{3})^{n - 1}

Apply the constant multiplication rule:

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

= 5 \cdot n = 1 \sum \infty (\frac{1}{3})^{n - 1}

Apply Series Ratio Test:

converges

= 5 converges

= converges

n = 0 \sum \infty \frac{3}{n !}

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

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

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

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

What is the sum from n=1 to infinity of 5(1/3)^{n-1} ?
The sum from n=1 to infinity of 5(1/3)^{n-1} is converges