What is the sum from n=0 to infinity of-6(1/4)^{2n} ?

The sum from n=0 to infinity of-6(1/4)^{2n} is converges

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sum from n=0 to infinity of-6(1/4)^{2n}

n = 0 \sum \infty - 6 (\frac{1}{4})^{2 n}

converges

Solution steps

n = 0 \sum \infty - 6 (\frac{1}{4})^{2 n}

Apply the constant multiplication rule:

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

= - 6 \cdot n = 0 \sum \infty (\frac{1}{4})^{2 n}

Apply Series Ratio Test:

converges

= - 6 converges

= converges

\int_{0}^{1} \frac{2 x}{( x ^{2} + 3 ) ^{3}} d x

x \to \infty lim \frac{( e ^{x^{4} - x^{2}} )}{2}

f (x) = x 1 + x^{2}

y^{^{'}} + \frac{2}{t} y = 0

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

What is the sum from n=0 to infinity of-6(1/4)^{2n} ?
The sum from n=0 to infinity of-6(1/4)^{2n} is converges