What is the sum from n=1 to infinity of (2^n)/(n2^n) ?

The sum from n=1 to infinity of (2^n)/(n2^n) is diverges

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sum from n=1 to infinity of (2^n)/(n2^n)

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

diverges

Solution steps

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

Cancel the common factor:

2^{n}

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

Apply Cauchy's Convergence Condition:

diverges

= diverges

n = 1 \sum \infty 0. 5^{n - 1}

n = 1 \sum \infty π^{\frac{1}{n}}

n = 0 \sum \infty \frac{4 ^{n}}{n}

n = 0 \sum \infty 4 - \frac{1}{n}

n = 3 \sum \infty \frac{1}{8 e ^{2 n}}

What is the sum from n=1 to infinity of (2^n)/(n2^n) ?
The sum from n=1 to infinity of (2^n)/(n2^n) is diverges