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

Lösung

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

1

Schritte zur Lösung

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

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

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

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

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

Vereinfache

= 1

n = 0 \sum \infty \frac{1}{8 - n}

n = 1 \sum \infty 9 (0.1)^{n - 1}

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

n = 1 \sum \infty 3 (0.3)^{n - 1}

n = 1 \sum \infty \frac{5 n ^{3}}{6 ^{n}}