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

Lösung

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

divergiert

Schritte zur Lösung

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

\frac{n + 1}{n} = (\frac{n}{n} + \frac{1}{n})

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

Vereinfache

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

Seriendivergenztest anwenden:

divergiert

= divergiert

k = 1 \sum \infty k e^{- 6 k^{2}}

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

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

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

n = 0 \sum \infty \frac{n}{1 + n}