What is the sum from n=1 to infinity}((-1)^{n-1 of n^2)/(12^n) ?

The sum from n=1 to infinity}((-1)^{n-1 of n^2)/(12^n) is converges

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sum from n=1 to infinity}((-1)^{n-1 of n^2)/(12^n)

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

converges

Solution steps

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

Apply Alternating Series Test:

converges

= converges

\frac{s + 3}{( s + 3 ) ^{2} + 9}

\frac{d}{d x} (x ln (- x))

x \to π lim (\frac{tan ( x )}{x})

\int_{7}^{\infty} x e^{- 5 x} d x

\frac{d}{d x} (4 x^{3} - 102 x^{2} + 840 x + 520)

What is the sum from n=1 to infinity}((-1)^{n-1 of n^2)/(12^n) ?
The sum from n=1 to infinity}((-1)^{n-1 of n^2)/(12^n) is converges