What is the sum from n=0 to infinity of (1+1/(2n))^n ?

The sum from n=0 to infinity of (1+1/(2n))^n is diverges

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sum from n=0 to infinity of (1+1/(2n))^n

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

diverges

Solution steps

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

\frac{1}{2 n} = \frac{1 + 2 n}{2 n} - 1

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

Simplify

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

Apply Series Divergence Test:

diverges

= diverges

n = 1 \sum \infty ln (\frac{3 n + 1}{n})

n = \frac{π}{2} \sum \infty sin (n)

n = 1 \sum \infty \frac{3}{( n ( n + 3 ) )}

k = 1 \sum \infty \frac{9}{k ( k + 2 )}

n = 1 \sum \infty \frac{- 5}{n}

What is the sum from n=0 to infinity of (1+1/(2n))^n ?
The sum from n=0 to infinity of (1+1/(2n))^n is diverges