What is the sum from n=5 to infinity of 0.9^{2(n+1)} ?

The sum from n=5 to infinity of 0.9^{2(n+1)} is converges

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sum from n=5 to infinity of 0.9^{2(n+1)}

n = 5 \sum \infty 0. 9^{2 (n + 1)}

converges

Solution steps

n = 5 \sum \infty 0. 9^{2 (n + 1)}

Simplify

0. 9^{2 (n + 1)} : 0.8 1^{n + 1}

= n = 5 \sum \infty 0.8 1^{n + 1}

Simplify

0.8 1^{n + 1} : 0.8 1^{n} \cdot 0.81

= n = 5 \sum \infty 0.8 1^{n} \cdot 0.81

Apply the constant multiplication rule:

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

= 0.81 \cdot n = 5 \sum \infty 0.8 1^{n}

Apply Series Ratio Test:

converges

= 0.81 converges

= converges

\int \frac{54}{x ^{2} - 81} d x

\int x^{3} ln (2 x) d x

\frac{d}{d x} (\frac{3 x + 1}{x - 1})

\frac{\partial}{\partial x} (e^{x} sin (x))

n \to \infty lim (\frac{n ^{2} + 3}{n ^{3} n + n ^{\frac{5}{2}}})

What is the sum from n=5 to infinity of 0.9^{2(n+1)} ?
The sum from n=5 to infinity of 0.9^{2(n+1)} is converges