What is the sum from n=1 to infinity of 5(-1/5)^n ?

The sum from n=1 to infinity of 5(-1/5)^n is converges

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sum from n=1 to infinity of 5(-1/5)^n

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Ratio Test:

converges

= 5 converges

= converges

f (x) = 2 x^{2} + x, at x = - 2

x \to 0 + lim (ln (x) \cdot ln (1 + x))

4 x^{\frac{5}{4}} + 8 x^{\frac{3}{2}} + 9 x

ln (x - 1)

\frac{d y}{d t} = \frac{e ^{y} + 1}{e ^{y}}

What is the sum from n=1 to infinity of 5(-1/5)^n ?
The sum from n=1 to infinity of 5(-1/5)^n is converges