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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Ratio Test:

converges

= 3 converges

= converges

n = 0 \sum \infty - \frac{3}{( n ^{2} - 4 )}

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

n = 0 \sum \infty \frac{1}{n ^{\frac{5}{2}}}

n = 0 \sum \infty \frac{n}{e ^{7 n}}

n = 4 \sum \infty \frac{5 ^{n}}{1 4 ^{n}}

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