sum from n=1 to infinity of 0.1^n

Lösung

n = 1 \sum \infty 0. 1^{n}

0.11111 \dots

Schritte zur Lösung

n = 1 \sum \infty 0. 1^{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. 1^{n} - 0. 1^{0}

n = 0 \sum \infty 0. 1^{n} = 1.11111 \dots

= 1.11111 \dots - 0. 1^{0}

Vereinfache

= 0.11111 \dots

i = 1 \sum \infty \frac{1}{i ( i + 1 )}

n = 0 \sum \infty \frac{6}{3 ^{n}}

n = 1 \sum \infty \frac{5}{3 n ^{3} + 1}

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

n = 1 \sum \infty 5 e^{- 0.5 n}