What is the sum from k=1 to infinity of 3(1/5)^{k-1} ?

The sum from k=1 to infinity of 3(1/5)^{k-1} is converges

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sum from k=1 to infinity of 3(1/5)^{k-1}

k = 1 \sum \infty 3 (\frac{1}{5})^{k - 1}

converges

Solution steps

k = 1 \sum \infty 3 (\frac{1}{5})^{k - 1}

Apply the constant multiplication rule:

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

= 3 \cdot k = 1 \sum \infty (\frac{1}{5})^{k - 1}

Apply Series Ratio Test:

converges

= 3 converges

= converges

\frac{\partial}{\partial y} (- 2)

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

x \to 0 lim (\frac{x - 1}{x ^{2}})

\frac{3}{x} + \frac{5}{x ^{3}}, at x = - 1

\frac{d}{d x} (\frac{x}{x - 3})

What is the sum from k=1 to infinity of 3(1/5)^{k-1} ?
The sum from k=1 to infinity of 3(1/5)^{k-1} is converges