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

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

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

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Divergence Test:

diverges

= 2 diverges

= diverges

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