What is the sum from n=0 to infinity of n*(3/4)^n ?

The sum from n=0 to infinity of n*(3/4)^n is converges

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sum from n=0 to infinity of n*(3/4)^n

n = 0 \sum \infty n \cdot (\frac{3}{4})^{n}

converges

Solution steps

n = 0 \sum \infty n (\frac{3}{4})^{n}

Apply Series Ratio Test:

converges

= converges

\int \frac{( x + 1 )}{x ^{3} + x ^{2} + 6 x} d x

x^{^{'}} = 4 x - x^{3}

\frac{d}{d x} (\frac{x - 5}{x + 6})

\frac{d}{d x} (b x)

y^{^{'}} = t + 3

What is the sum from n=0 to infinity of n*(3/4)^n ?
The sum from n=0 to infinity of n*(3/4)^n is converges