What is the sum from n=1 to infinity of 1/n (4/5)^n ?

The sum from n=1 to infinity of 1/n (4/5)^n is converges

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sum from n=1 to infinity of 1/n (4/5)^n

n = 1 \sum \infty \frac{1}{n} (\frac{4}{5})^{n}

converges

Solution steps

n = 1 \sum \infty \frac{1}{n} (\frac{4}{5})^{n}

Apply Series Ratio Test:

converges

= converges

n = 1 \sum \infty cos (\frac{9}{n})

n = 3 \sum \infty ln (1 + \frac{1}{n})

n = 2 \sum \infty \frac{6}{n ln ( n )}

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

n = 1 \sum \infty n (- \frac{5}{9})^{n}

What is the sum from n=1 to infinity of 1/n (4/5)^n ?
The sum from n=1 to infinity of 1/n (4/5)^n is converges