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

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

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

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

diverges

Solution steps

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

(- 2)^{n} = 2^{n} (- 1)^{n}

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

Apply Series Ratio Test:

diverges

= diverges

n = 1 \sum \infty \frac{n !}{6 ^{n}}

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

n = 0 \sum \infty \frac{n}{10 n + 1}

n = 0 \sum \infty n ln (n)

n = 0 \sum \infty \frac{2}{1 0 ^{n}}

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