What is the sum from n=0 to infinity of n*(4/5)^n ?

The sum from n=0 to infinity of n*(4/5)^n is converges

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sum from n=0 to infinity of n*(4/5)^n

n = 0 \sum \infty n \cdot (\frac{4}{5})^{n}

converges

Solution steps

n = 0 \sum \infty n (\frac{4}{5})^{n}

Apply Series Ratio Test:

converges

= converges

y = \frac{529 - 8 x}{7 + 7 x}, (4, 7)

\frac{d}{d x} (\frac{a x ^{2} + b x + 1}{x ^{2} - 6 x + 8})

\frac{d ^{2}}{d x ^{2}} ((3 + 2 x) e^{- 5 x})

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

x \to 2 lim (3 x^{2} - 5 x + 1)

What is the sum from n=0 to infinity of n*(4/5)^n ?
The sum from n=0 to infinity of n*(4/5)^n is converges