sum from n=2 to infinity of 0.33^n

Lösung

n = 2 \sum \infty 0.3 3^{n}

0.16253 \dots

Schritte zur Lösung

n = 2 \sum \infty 0.3 3^{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 0.3 3^{n} - 0.3 3^{0} - 0.3 3^{1}

n = 0 \sum \infty 0.3 3^{n} = 1.49253 \dots

= 1.49253 \dots - 0.3 3^{0} - 0.3 3^{1}

Vereinfache

= 0.16253 \dots

n = 3 \sum \infty \frac{7 ^{n}}{1 0 ^{n}}

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

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

n = 0 \sum \infty \frac{6}{n}

i = 1 \sum \infty e^{- i}