What is the sum from n=1 to infinity of ((2-3)^n)/n ?

The sum from n=1 to infinity of ((2-3)^n)/n is converges

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sum from n=1 to infinity of ((2-3)^n)/n

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

converges

Solution steps

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

Simplify

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

Apply Alternating Series Test:

converges

= converges

n = 1 \sum \infty \frac{1}{n ^{2} + 7 n + 1}

k = 0 \sum \infty (- \frac{1}{2})^{k}

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

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

k = 0 \sum \infty (\frac{3}{4})^{k}

What is the sum from n=1 to infinity of ((2-3)^n)/n ?
The sum from n=1 to infinity of ((2-3)^n)/n is converges