sum from m=2 to infinity of 6/(7^m)

Lösung

m = 2 \sum \infty \frac{6}{7 ^{m}}

\frac{1}{7}

Dezimale

0.14285 \dots

Schritte zur Lösung

m = 2 \sum \infty \frac{6}{7 ^{m}}

Wende die Regel der konstanten Multiplikation:

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

= 6 \cdot m = 2 \sum \infty \frac{1}{7 ^{m}}

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

= 6 (m = 0 \sum \infty \frac{1}{7 ^{m}} - \frac{1}{7 ^{0}} - \frac{1}{7 ^{1}})

m = 0 \sum \infty \frac{1}{7 ^{m}} = \frac{7}{6}

= 6 (\frac{7}{6} - \frac{1}{7 ^{0}} - \frac{1}{7 ^{1}})

Vereinfache

6 (\frac{7}{6} - \frac{1}{7 ^{0}} - \frac{1}{7 ^{1}}) : \frac{1}{7}

= \frac{1}{7}

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

k = 1 \sum \infty 1^{k}

n = 0 \sum \infty 2021

n = 1 \sum \infty (- 1)^{n} \frac{3 n}{7 n ^{2} + 1}

n = 0 \sum \infty \frac{4}{n ^{2} - 4}