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

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

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

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

converges

Solution steps

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

Apply Alternating Series Test:

converges

= converges

\int (- 21 x^{2} - 4 x) cos (- 7 x^{3} - 2 x^{2} - 2) d x

f (x) = 2 x^{2}, at x = 2

\frac{\partial}{\partial x} (4 x^{3} y^{2} - 3)

\int - 2 + 24 x - 12 x^{2} d x

x x^{2} - 1

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