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

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

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

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

converges

Solution steps

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

Simplify

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

Apply Alternating Series Test:

converges

= converges

n = 0 \sum \infty \frac{2}{1 + 3 n}

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

n = 0 \sum \infty 8 + 4 + 2 + 1

n = 1 \sum \infty (- \frac{9}{8})^{n - 1}

n = 0 \sum \infty \frac{1}{1 + 3 n ^{4}}

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