What is the sum from n=1 to infinity of n^2*(0.5)^n ?

The sum from n=1 to infinity of n^2*(0.5)^n is converges

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sum from n=1 to infinity of n^2*(0.5)^n

n = 1 \sum \infty n^{2} \cdot (0.5)^{n}

converges

Solution steps

n = 1 \sum \infty n^{2} \cdot 0. 5^{n}

Apply Series Ratio Test:

converges

= converges

\int_{1}^{\infty} 14 e^{- 7 x} d x

\frac{d y}{d x} = - 8 x y^{2}

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

n = 1 \sum \infty (\frac{- 1}{2})^{n}

\frac{\partial}{\partial x} ((4 x + 4 y) e^{y})

What is the sum from n=1 to infinity of n^2*(0.5)^n ?
The sum from n=1 to infinity of n^2*(0.5)^n is converges