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Beliebt Rechnen >

sum from n=1 to infinity of 4^{-n}

Lösung

n = 1 \sum \infty 4^{- n}

Lösung

\frac{1}{3}

+1

Dezimale

0.33333 \dots

Schritte zur Lösung

n = 1 \sum \infty 4^{- n}

Vereinfache

4^{- n} : \frac{1}{4 ^{n}}

= n = 1 \sum \infty \frac{1}{4 ^{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}{4 ^{n}} - \frac{1}{4 ^{0}}

n = 0 \sum \infty \frac{1}{4 ^{n}} = \frac{4}{3}

= \frac{4}{3} - \frac{1}{4 ^{0}}

Vereinfache

\frac{4}{3} - \frac{1}{4 ^{0}} : \frac{1}{3}

= \frac{1}{3}

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