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

Lösung

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

divergiert

Schritte zur Lösung

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}

Vereinfache

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

Seriendivergenztest anwenden:

divergiert

= divergiert

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

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