sum from n=1 to infinity of 90(0.1)^n

Lösung

n = 1 \sum \infty 90 (0.1)^{n}

10

Schritte zur Lösung

n = 1 \sum \infty 90 \cdot 0. 1^{n}

Wende die Regel der konstanten Multiplikation:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 90 \cdot 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}

= 90 (n = 0 \sum \infty 0. 1^{n} - 0. 1^{0})

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

= 90 (1.11111 \dots - 0. 1^{0})

Vereinfache

= 10

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

m = 1 \sum \infty \frac{10}{m ^{8} + 3}

n = 1 \sum \infty \frac{n}{n ^{2} + 6}

k = 1 \sum \infty \frac{5}{k} - \frac{5}{k + 1}

n = 1 \sum \infty \frac{5}{n ( n + 1 )}