What is the sum from n=1 to infinity}((-1)^{n-1 of n^2)/(10^n) ?

The sum from n=1 to infinity}((-1)^{n-1 of n^2)/(10^n) is converges

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sum from n=1 to infinity}((-1)^{n-1 of n^2)/(10^n)

n = 1 \sum \infty \frac{( - 1 ) ^{n - 1} n ^{2}}{1 0 ^{n}}

converges

Solution steps

n = 1 \sum \infty \frac{( - 1 ) ^{n - 1} n ^{2}}{1 0 ^{n}}

Apply Alternating Series Test:

converges

= converges

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

\frac{d}{d x} (sin (3 x^{4}))

\frac{d}{d x} (6 sin (x) - 8 e^{x})

\frac{d ^{2}}{d x ^{2}} (\frac{1 - 5 x}{1 + 5 x})

f (x) = 2 x^{2} - 5 x, (- 1, 7)

What is the sum from n=1 to infinity}((-1)^{n-1 of n^2)/(10^n) ?
The sum from n=1 to infinity}((-1)^{n-1 of n^2)/(10^n) is converges