sum from n=2 to infinity of 4/(5^n)

Lösung

n = 2 \sum \infty \frac{4}{5 ^{n}}

\frac{1}{5}

Dezimale

0.2

Schritte zur Lösung

n = 2 \sum \infty \frac{4}{5 ^{n}}

Wende die Regel der konstanten Multiplikation:

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

= 4 \cdot n = 2 \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}

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

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

= 4 (\frac{5}{4} - \frac{1}{5 ^{0}} - \frac{1}{5 ^{1}})

Vereinfache

4 (\frac{5}{4} - \frac{1}{5 ^{0}} - \frac{1}{5 ^{1}}) : \frac{1}{5}

= \frac{1}{5}

n = 0 \sum \infty \frac{5}{1 + 9 n ^{2}}

n = 0 \sum \infty ln (4 n + 1)

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

n = 0 \sum \infty \frac{5}{n + 1}

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