sum from n=2 to infinity of 1/(2^{n-1)}

Lösung

n = 2 \sum \infty \frac{1}{2 ^{n - 1}}

1

Schritte zur Lösung

n = 2 \sum \infty \frac{1}{2 ^{n - 1}}

Vereinfache

2^{n - 1} : 2^{n} \cdot 2^{- 1}

= n = 2 \sum \infty \frac{1}{2 ^{n} \cdot 2 ^{- 1}}

Wende die Regel der konstanten Multiplikation:

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

= \frac{1}{2 ^{- 1}} \cdot n = 2 \sum \infty \frac{1}{2 ^{n}}

\frac{1}{2 ^{- 1}} = 2

= 2 \cdot n = 2 \sum \infty \frac{1}{2 ^{n}}

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

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

n = 0 \sum \infty \frac{1}{2 ^{n}} = 2

= 2 (2 - \frac{1}{2 ^{0}} - \frac{1}{2 ^{1}})

Vereinfache

2 (2 - \frac{1}{2 ^{0}} - \frac{1}{2 ^{1}}) : 1

= 1

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