What is the sum from n=2 to infinity of n*(3/4)^n ?

The sum from n=2 to infinity of n*(3/4)^n is converges

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sum from n=2 to infinity of n*(3/4)^n

n = 2 \sum \infty n \cdot (\frac{3}{4})^{n}

converges

Solution steps

n = 2 \sum \infty n (\frac{3}{4})^{n}

Apply Series Ratio Test:

converges

= converges

y^{^{''}} = 6 x + 2, y (1) = - 1, y^{^{'}} (1) = 7

\frac{\partial}{\partial z} (arctan (x^{2} y z^{2}))

\frac{\partial}{\partial z} (x arctan (yz))

\frac{\partial}{\partial x} ((2 x + 3 y)^{2})

x \to 0 lim (\frac{1}{x - 3})

What is the sum from n=2 to infinity of n*(3/4)^n ?
The sum from n=2 to infinity of n*(3/4)^n is converges