What is the sum from n=1 to infinity of (-1/4)^{n-1} ?

The sum from n=1 to infinity of (-1/4)^{n-1} is converges

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sum from n=1 to infinity of (-1/4)^{n-1}

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

n = 0 \sum \infty \frac{6}{7 ^{n}}

x = 1 \sum \infty \frac{2 ^{x}}{5 ^{x}}

n = 0 \sum \infty cos (6 n)

n = 0 \sum \infty (\frac{1}{23})^{n}

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

What is the sum from n=1 to infinity of (-1/4)^{n-1} ?
The sum from n=1 to infinity of (-1/4)^{n-1} is converges