sum from n=1 to infinity of (1/100)^n

Lösung

n = 1 \sum \infty (\frac{1}{100})^{n}

\frac{1}{99}

Dezimale

0.01010 \dots

Schritte zur Lösung

n = 1 \sum \infty (\frac{1}{100})^{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}{100})^{n} - (\frac{1}{100})^{0}

n = 0 \sum \infty (\frac{1}{100})^{n} = \frac{100}{99}

= \frac{100}{99} - (\frac{1}{100})^{0}

Vereinfache

\frac{100}{99} - (\frac{1}{100})^{0} : \frac{1}{99}

= \frac{1}{99}

n = 0 \sum \infty \frac{n}{n ^{3}}

n = 6 \sum \infty ln (\frac{n}{3})

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

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

n = 1 \sum \infty \frac{4}{n ^{- 2.3}}