What is the sum from n=0 to infinity of (n^2)/(2^n) ?

The sum from n=0 to infinity of (n^2)/(2^n) is converges

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sum from n=0 to infinity of (n^2)/(2^n)

n = 0 \sum \infty \frac{n ^{2}}{2 ^{n}}

converges

Solution steps

n = 0 \sum \infty \frac{n ^{2}}{2 ^{n}}

Apply Series Ratio Test:

converges

= converges

\frac{d}{d x} (x sin (\frac{6}{x}))

\int \frac{x}{( a + b x ^{2} ) ^{3}} d x

y = \frac{x ^{2}}{2} - 3, y = - \frac{x ^{2}}{2} + 2 x

\int_{1}^{9} \frac{1}{x - 7} d x

\frac{d}{d x} (- x e^{(\frac{x + 1}{x - 1})})

What is the sum from n=0 to infinity of (n^2)/(2^n) ?
The sum from n=0 to infinity of (n^2)/(2^n) is converges