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

The sum from n=1 to infinity of ((2)^n)/n is diverges

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

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

diverges

Solution steps

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

Apply Series Ratio Test:

diverges

= diverges

k = 4 \sum \infty \frac{6}{k} - \frac{6}{k + 1}

n = 0 \sum \infty (\frac{3}{n})^{\frac{3}{n}}

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

n = 0 \sum \infty 4 (\frac{8}{9})^{n}

n = 1 \sum \infty cos (n)

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