What is the sum from n=0 to infinity of 7(-1/7)^n ?

The sum from n=0 to infinity of 7(-1/7)^n is 49/8

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sum from n=0 to infinity of 7(-1/7)^n

n = 0 \sum \infty 7 (- \frac{1}{7})^{n}

\frac{49}{8}

Solution steps

n = 0 \sum \infty 7 (- \frac{1}{7})^{n}

Apply the constant multiplication rule:

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

= 7 \cdot n = 0 \sum \infty (- \frac{1}{7})^{n}

Apply Series Geometric Test:

\frac{7}{8}

= 7 \cdot \frac{7}{8}

Simplify

7 \cdot \frac{7}{8} : \frac{49}{8}

= \frac{49}{8}

\frac{\partial}{\partial x} (- x sin (y) + e^{x})

y^{^{''}} - 4 y^{^{'}} + 2 y = 0, y (0) = 0, y^{^{'}} (0) = 7

x \to \infty lim (\frac{1}{8})

f (x) = \frac{3}{4 x + 1}, (- 1, - 1)

f (x) = \frac{( x ^{2} + 1 )}{( x ^{2} - 1 )}

What is the sum from n=0 to infinity of 7(-1/7)^n ?
The sum from n=0 to infinity of 7(-1/7)^n is 49/8