sum from n=1 to infinity of (6/16)^n

Lösung

n = 1 \sum \infty (\frac{6}{16})^{n}

\frac{3}{5}

Dezimale

0.6

Schritte zur Lösung

n = 1 \sum \infty (\frac{6}{16})^{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{6}{16})^{n} - (\frac{6}{16})^{0}

n = 0 \sum \infty (\frac{6}{16})^{n} = \frac{8}{5}

= \frac{8}{5} - (\frac{6}{16})^{0}

Vereinfache

\frac{8}{5} - (\frac{6}{16})^{0} : \frac{3}{5}

= \frac{3}{5}

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

n = 0 \sum \infty \frac{1000}{n}

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

n = 0 \sum \infty \frac{e}{n}

n = 0 \sum \infty \frac{7}{π ^{n}}