sum from n=1 to infinity of 2/(e^{n-3)}

Lösung

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

2 e^{3} (\frac{e}{e - 1} - 1)

Dezimale

23.37862 \dots

Schritte zur Lösung

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

Wende die Regel der konstanten Multiplikation:

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

= 2 \cdot n = 1 \sum \infty \frac{1}{e ^{n - 3}}

Vereinfache

e^{n - 3} : e^{n} e^{- 3}

= 2 \cdot n = 1 \sum \infty \frac{1}{e ^{n} e ^{- 3}}

Wende die Regel der konstanten Multiplikation:

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

= 2 \cdot \frac{1}{e ^{- 3}} \cdot n = 1 \sum \infty \frac{1}{e ^{n}}

\frac{1}{e ^{- 3}} = e^{3}

= 2 e^{3} \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}

= 2 e^{3} (n = 0 \sum \infty \frac{1}{e ^{n}} - \frac{1}{e ^{0}})

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

= 2 e^{3} (\frac{e}{e - 1} - \frac{1}{e ^{0}})

Vereinfache

= 2 e^{3} (\frac{e}{e - 1} - 1)

y^{^{''}} - \frac{2}{x} y^{^{'}} - \frac{4}{x ^{2}} y = 3 x^{2}

x \to 1 + lim (x^{3} - 4)

x \to - 1 lim (\frac{2 x}{x + 1})

n = 1 \sum \infty \frac{3}{n ^{4}}

h \to 0 lim (\frac{( 1 + h ) ^{2} - 1}{h})