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

Lösung

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

5 (\frac{e}{e - 1} - 1)

Dezimale

2.90988 \dots

Schritte zur Lösung

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

Wende die Regel der konstanten Multiplikation:

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

= 5 \cdot n = 1 \sum \infty \frac{1}{e ^{n}}

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

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

n = 0 \sum \infty \frac{1}{e ^{n}} = \frac{e}{e - 1}

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

Vereinfache

= 5 (\frac{e}{e - 1} - 1)