What is the sum from n=5 to infinity of 3(0.9)^{n+1} ?

The sum from n=5 to infinity of 3(0.9)^{n+1} is converges

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sum from n=5 to infinity of 3(0.9)^{n+1}

n = 5 \sum \infty 3 (0.9)^{n + 1}

converges

Solution steps

n = 5 \sum \infty 3 \cdot 0. 9^{n + 1}

Apply the constant multiplication rule:

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

= 3 \cdot n = 5 \sum \infty 0. 9^{n + 1}

Simplify

0. 9^{n + 1} : 0. 9^{n} \cdot 0.9

= 3 \cdot n = 5 \sum \infty 0. 9^{n} \cdot 0.9

Apply the constant multiplication rule:

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

= 3 \cdot 0.9 \cdot n = 5 \sum \infty 0. 9^{n}

Apply Series Ratio Test:

converges

= 3 \cdot 0.9 converges

= converges

x \to 0 lim (x - ln (x))

y^{^{'}} + y^{5} x + 9 y = 0

x \to - \frac{π}{2} lim (tan (x))

\int \frac{1}{2 e ^{2 x}} d x

What is the sum from n=5 to infinity of 3(0.9)^{n+1} ?
The sum from n=5 to infinity of 3(0.9)^{n+1} is converges