What is the sum from n=1 to infinity of 4(-10/11)^n ?

The sum from n=1 to infinity of 4(-10/11)^n is converges

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sum from n=1 to infinity of 4(-10/11)^n

n = 1 \sum \infty 4 (- \frac{10}{11})^{n}

converges

Solution steps

n = 1 \sum \infty 4 (- \frac{10}{11})^{n}

Apply the constant multiplication rule:

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

= 4 \cdot n = 1 \sum \infty (- \frac{10}{11})^{n}

Apply Series Ratio Test:

converges

= 4 converges

= converges

k = 1 \sum \infty \frac{k}{e ^{4 k}}

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

n = 5 \sum \infty \frac{1}{n ^{2} - 5 n + 6}

n = 1 \sum \infty \frac{1}{n ^{\frac{5}{6}}}

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

What is the sum from n=1 to infinity of 4(-10/11)^n ?
The sum from n=1 to infinity of 4(-10/11)^n is converges