What is the sum from n=1 to infinity of 8(-1/4)^n ?

The sum from n=1 to infinity of 8(-1/4)^n is converges

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sum from n=1 to infinity of 8(-1/4)^n

n = 1 \sum \infty 8 (- \frac{1}{4})^{n}

converges

Solution steps

n = 1 \sum \infty 8 (- \frac{1}{4})^{n}

Apply the constant multiplication rule:

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

= 8 \cdot n = 1 \sum \infty (- \frac{1}{4})^{n}

Apply Series Ratio Test:

converges

= 8 converges

= converges

x \to 0 + lim (2 x ln (x))

x^{2} + 4, - x^{2} - 1, x = - 2, x = 3

f (x) = x^{2} ln (2 x)

x y^{^{'}} = y + 5 x^{2} sin (x)

f (x) = (2 x^{3} + 6 x) (4 x^{2} - 7 x + 11)

What is the sum from n=1 to infinity of 8(-1/4)^n ?
The sum from n=1 to infinity of 8(-1/4)^n is converges