sum from n=1 to infinity of 9/(10^n)

Lösung

n = 1 \sum \infty \frac{9}{1 0 ^{n}}

1

Schritte zur Lösung

n = 1 \sum \infty \frac{9}{1 0 ^{n}}

Wende die Regel der konstanten Multiplikation:

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

= 9 \cdot n = 1 \sum \infty \frac{1}{1 0 ^{n}}

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

= 9 (n = 0 \sum \infty \frac{1}{1 0 ^{n}} - \frac{1}{1 0 ^{0}})

n = 0 \sum \infty \frac{1}{1 0 ^{n}} = \frac{10}{9}

= 9 (\frac{10}{9} - \frac{1}{1 0 ^{0}})

Vereinfache

9 (\frac{10}{9} - \frac{1}{1 0 ^{0}}) : 1

= 1

\frac{d}{d t} (cos (3 t))

n = 0 \sum \infty \frac{1}{n \cdot ln ( n )}

\int e^{- 5 x} tan (e^{- 5 x}) d x

\frac{d}{d x} (1 + sec (x))

\frac{d}{d x} (- x^{5} - 6 x^{4} - 14 x^{3} - 16 x^{2} - 9 x - 2)