What is the sum from n=1 to infinity of (-1)^n 5/n ?

The sum from n=1 to infinity of (-1)^n 5/n is converges

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sum from n=1 to infinity of (-1)^n 5/n

n = 1 \sum \infty (- 1)^{n} \frac{5}{n}

converges

Solution steps

n = 1 \sum \infty (- 1)^{n} \frac{5}{n}

Apply the constant multiplication rule:

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

= 5 \cdot n = 1 \sum \infty (- 1)^{n} \frac{1}{n}

Apply Alternating Series Test:

converges

= 5 converges

= converges

n = 0 \sum \infty 10 (- \frac{6}{10})^{n}

n = 1 \sum \infty \frac{1}{3 ^{n} - 2 ^{n}}

n = 2 \sum \infty n e^{- n^{2}}

n = 1 \sum \infty \frac{n ^{3}}{n + 1}

n = 2 \sum \infty \frac{7}{5 ^{n}}

What is the sum from n=1 to infinity of (-1)^n 5/n ?
The sum from n=1 to infinity of (-1)^n 5/n is converges