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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Alternating Series Test:

converges

= 2 converges

= converges

n = 1 \sum \infty \frac{6 n}{4 n + 1}

n = 1 \sum \infty \frac{2}{5 n ^{2} + 5 n}

n = 0 \sum \infty (\frac{5}{12})^{n}

n = 0 \sum \infty \frac{3}{5 ^{n}} + \frac{2}{n}

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

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