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

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

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

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

converges

Solution steps

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

Apply Series Integral Test:

converges

= converges

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

n = 0 \sum \infty n 2^{- n}

n = 1 \sum \infty 9 (0.4)^{n - 1}

n = 1 \sum \infty \frac{2}{n ^{2} + 9}

k = 1 \sum \infty \frac{k ^{2}}{3 ^{k}}

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