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

Lösung

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

\frac{1}{9}

Dezimale

0.11111 \dots

Schritte zur Lösung

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}

= 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}

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

Vereinfache

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

= \frac{1}{9}

n = 1 \sum \infty (\frac{2}{7}) \cdot n

\int 2 t + 2 d t

n = 0 \sum \infty \frac{- 14}{8}

\frac{d}{d x} (5 ln (x))

\frac{d}{d x} (3 sin (x) + 2 cos (x) tan (x))