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

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

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

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

diverges

Solution steps

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

Apply Series Ratio Test:

diverges

= diverges

k = 2 \sum \infty (- 0.35)^{k}

n = 1 \sum \infty \frac{7 ^{n}}{1 0 ^{n}}

k = 5 \sum \infty 2 (- \frac{1}{4})^{k}

n = 4 \sum \infty 4 (6 - 2 n)

n = 1 \sum \infty 3^{n} 4^{- n + 1}

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