What is the sum from k=4 to infinity of 3(-1/2)^k ?

The sum from k=4 to infinity of 3(-1/2)^k is converges

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sum from k=4 to infinity of 3(-1/2)^k

k = 4 \sum \infty 3 (- \frac{1}{2})^{k}

converges

Solution steps

k = 4 \sum \infty 3 (- \frac{1}{2})^{k}

Apply the constant multiplication rule:

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

= 3 \cdot k = 4 \sum \infty (- \frac{1}{2})^{k}

Apply Series Ratio Test:

converges

= 3 converges

= converges

\int e^{2 x} sin (5 x) d x

y = 55 cos (0.071 x + 0.39) + 63

\int \frac{e ^{3 x} + 2 e ^{x}}{e ^{3 x} + 1} d x

\frac{\partial}{\partial x} (5 y e^{4 x})

(- 2, 3), (2, - 2)

What is the sum from k=4 to infinity of 3(-1/2)^k ?
The sum from k=4 to infinity of 3(-1/2)^k is converges