sum from n=1 to infinity of 2/(4^n)

Lösung

n = 1 \sum \infty \frac{2}{4 ^{n}}

\frac{2}{3}

Dezimale

0.66666 \dots

Schritte zur Lösung

n = 1 \sum \infty \frac{2}{4 ^{n}}

Wende die Regel der konstanten Multiplikation:

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

= 2 \cdot 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}

= 2 (n = 0 \sum \infty \frac{1}{4 ^{n}} - \frac{1}{4 ^{0}})

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

= 2 (\frac{4}{3} - \frac{1}{4 ^{0}})

Vereinfache

2 (\frac{4}{3} - \frac{1}{4 ^{0}}) : \frac{2}{3}

= \frac{2}{3}

n = 0 \sum \infty \frac{6 ^{n}}{n !}

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

n = 1 \sum \infty \frac{1}{2 ^{n} + 3 n}

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

n = 0 \sum \infty \frac{3}{2 n}