What is the sum from n=1 to infinity of ((0.99)^n)/n ?

The sum from n=1 to infinity of ((0.99)^n)/n is converges

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sum from n=1 to infinity of ((0.99)^n)/n

n = 1 \sum \infty \frac{( 0.99 ) ^{n}}{n}

converges

Solution steps

n = 1 \sum \infty \frac{0.9 9 ^{n}}{n}

Apply Series Ratio Test:

converges

= converges

k = 0 \sum \infty 3 (- \frac{1}{4})^{k}

n = 1 \sum \infty \frac{3}{5 n + 1}

i = 1 \sum \infty 7 + 3 i

n = 1 \sum \infty (- 0.45)^{n}

k = 1 \sum \infty \frac{k}{k + 1}

What is the sum from n=1 to infinity of ((0.99)^n)/n ?
The sum from n=1 to infinity of ((0.99)^n)/n is converges