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

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

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

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

diverges

Solution steps

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

(1 - 3)^{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{1}{10})^{n - 1}

n = 1 \sum \infty 5 (\frac{3}{7})^{n}

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

n = 1 \sum \infty \frac{1}{24} (\frac{3}{4})^{n}

n = 0 \sum \infty A n \cdot (- 1)^{n}

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