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

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

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

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

converges

Solution steps

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

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

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

n^{- 1} = \frac{1}{n}

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

Apply Series Ratio Test:

converges

= converges

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

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

n = 0 \sum \infty (\frac{1}{n})^{n}

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

n = 1 \sum \infty \frac{1}{n ^{0.8}}

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