What is the sum from i=1 to infinity of 3(1/4)^{i-1} ?

The sum from i=1 to infinity of 3(1/4)^{i-1} is converges

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sum from i=1 to infinity of 3(1/4)^{i-1}

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Ratio Test:

converges

= 3 converges

= converges

\int_{0}^{1} e^{x} tan (e^{x}) d x

x \to \infty + lim (3^{- x})

x \to 1 lim (\frac{in ( x )}{x - 1})

y = x^{2} + 10 x

\frac{d y}{d x} = \frac{x ^{2} - 2 y}{x}, y (2) = 2

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