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

Lösung

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

\frac{1}{4}

Dezimale

0.25

Schritte zur Lösung

n = 3 \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}

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

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

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

Vereinfache

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

= \frac{1}{4}

n = 1 \sum \infty \frac{1}{( 6 n ) ^{6}}

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

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

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

n = 1 \sum \infty (\frac{7 π}{8})^{n}