sum from n=2 to infinity of 1/(3(2^n))

Lösung

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

\frac{1}{6}

Dezimale

0.16666 \dots

Schritte zur Lösung

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

Wende die Regel der konstanten Multiplikation:

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

= \frac{1}{3} \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}

= \frac{1}{3} (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

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

Vereinfache

\frac{1}{3} (2 - \frac{1}{2 ^{0}} - \frac{1}{2 ^{1}}) : \frac{1}{6}

= \frac{1}{6}

n = 70 \sum \infty \frac{3}{n ^{2} + 2 n}

k = 1 \sum \infty (- \frac{1}{3})^{k}

n = 1 \sum \infty (\frac{3}{5})^{n - 1}

n = 0 \sum \infty \frac{5}{5 - n}

n = 1 \sum \infty \frac{1}{n ^{3} + 6}