sum from n=3 to infinity of 1/(5^n)

Lösung

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

\frac{1}{100}

Dezimale

0.01

Schritte zur Lösung

n = 3 \sum \infty \frac{1}{5 ^{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}{5 ^{n}} - \frac{1}{5 ^{0}} - \frac{1}{5 ^{1}} - \frac{1}{5 ^{2}}

n = 0 \sum \infty \frac{1}{5 ^{n}} = \frac{5}{4}

= \frac{5}{4} - \frac{1}{5 ^{0}} - \frac{1}{5 ^{1}} - \frac{1}{5 ^{2}}

Vereinfache

\frac{5}{4} - \frac{1}{5 ^{0}} - \frac{1}{5 ^{1}} - \frac{1}{5 ^{2}} : \frac{1}{100}

= \frac{1}{100}

n = 0 \sum \infty - (\frac{7}{3 ^{n}})

n = 1 \sum \infty \frac{8}{- 3 ^{n}}

n = 2 \sum \infty \frac{- 2}{n + 1}

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

n = 0 \sum \infty 8 ln (4 - n)