What is the sum from n=0 to infinity of 5(0.7)^{n-1} ?

The sum from n=0 to infinity of 5(0.7)^{n-1} is converges

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sum from n=0 to infinity of 5(0.7)^{n-1}

n = 0 \sum \infty 5 (0.7)^{n - 1}

converges

Solution steps

n = 0 \sum \infty 5 \cdot 0. 7^{n - 1}

Apply the constant multiplication rule:

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

= 5 \cdot n = 0 \sum \infty 0. 7^{n - 1}

Apply Series Ratio Test:

converges

= 5 converges

= converges

\int e^{4 x} d x

\int \frac{( - 1 ) ^{k}}{( 2 k + 1 ) !} (2 x)^{2 k} d x

\frac{d x}{d t} = x + t

\frac{3}{( 45 + 36 ) ^{3}}

e^{x^{7}}

What is the sum from n=0 to infinity of 5(0.7)^{n-1} ?
The sum from n=0 to infinity of 5(0.7)^{n-1} is converges