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

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

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sum from n=1 to infinity of-5/(2^{n-1)}

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Ratio Test:

converges

= - 5 converges

= converges

x \to \infty lim (3)

\int (x + π) d x

F (s) = \frac{( s + 2 ) ^{2}}{s ^{3}}

y = (x^{2} - 8 x) e^{x}

f (x) = \frac{2 x - 4}{x + 1}

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