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

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

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

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

converges

Solution steps

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

Apply Alternating Series Test:

converges

= converges

\int \frac{ln ( x ^{\frac{7}{2}} )}{x} d x

\frac{d}{d x} ((ln (x))^{p (x)})

\frac{d y}{d x} = \frac{y}{( x ^{2} + 1 )}

\frac{16}{s ( s + 2 ) ^{4}}

x \to \infty lim (\frac{e ^{- s x}}{s})

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