sum from n=1 to infinity of-1/(18^n)

Lösung

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

- \frac{1}{17}

Dezimale

- 0.05882 \dots

Schritte zur Lösung

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

Wende die Regel der konstanten Multiplikation:

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

= - n = 1 \sum \infty \frac{1}{1 8 ^{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 \frac{1}{1 8 ^{n}} - \frac{1}{1 8 ^{0}})

n = 0 \sum \infty \frac{1}{1 8 ^{n}} = \frac{18}{17}

= - (\frac{18}{17} - \frac{1}{1 8 ^{0}})

Vereinfache

- (\frac{18}{17} - \frac{1}{1 8 ^{0}}) : - \frac{1}{17}

= - \frac{1}{17}

n = 1 \sum \infty (cos (14))^{n}

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

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

n = 0 \sum \infty (\frac{5}{2})^{n - 1}

n = 1 \sum \infty n^{3} \frac{1}{n !}