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

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

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sum from n=1 to infinity}((2^{-n of))/n

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

n = 0 \sum \infty n \cdot (\frac{1}{2})^{n}

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

n = 0 \sum \infty 10 (- \frac{6}{10})^{n}

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

n = 2 \sum \infty n e^{- n^{2}}

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