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

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

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

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

1

Solution steps

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

Cancel the common factor:

n

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

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

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

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

= 2 - (\frac{1}{2})^{0}

Simplify

= 1

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