sum from n=1 to infinity of 3(2/5)^n

Lösung

n = 1 \sum \infty 3 (\frac{2}{5})^{n}

2

Schritte zur Lösung

n = 1 \sum \infty 3 (\frac{2}{5})^{n}

Wende die Regel der konstanten Multiplikation:

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

= 3 \cdot n = 1 \sum \infty (\frac{2}{5})^{n}

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

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

n = 0 \sum \infty (\frac{2}{5})^{n} = \frac{5}{3}

= 3 (\frac{5}{3} - (\frac{2}{5})^{0})

Vereinfache

3 (\frac{5}{3} - (\frac{2}{5})^{0}) : 2

= 2

n = 1 \sum \infty \frac{ln ( n ^{4} )}{n}

n = 2 \sum \infty \frac{5}{n ( n + 2 )}

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

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

n = 0 \sum \infty e^{- 19 n}