sum from n=1 to infinity of 70(0.1)^n

Lösung

n = 1 \sum \infty 70 (0.1)^{n}

7.77777 \dots

Schritte zur Lösung

n = 1 \sum \infty 70 \cdot 0. 1^{n}

Wende die Regel der konstanten Multiplikation:

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

= 70 \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}

= 70 (n = 0 \sum \infty 0. 1^{n} - 0. 1^{0})

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

= 70 (1.11111 \dots - 0. 1^{0})

Vereinfache

= 7.77777 \dots

n = 1 \sum \infty \frac{8 ^{n}}{7 ^{n}}

n = 6 \sum \infty ln (1 + \frac{1}{n})

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

n = 1 \sum \infty 5 \cdot 0. 2^{n - 1}

k = 1 \sum \infty \frac{2}{k ( k + 2 )}