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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

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

Apply Alternating Series Test:

converges

= 2 converges

= converges

y = 8 x e^{x}

\frac{d}{d z} (\frac{2 z ^{6} + 2}{1 z ^{2} + 5})

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

\int cos^{4} (1) d x

\int 10 x ln (x + 5) d x

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