sum from n=1 to infinity of (2^n)/(4^n)

Lösung

n = 1 \sum \infty \frac{2 ^{n}}{4 ^{n}}

1

Schritte zur Lösung

n = 1 \sum \infty \frac{2 ^{n}}{4 ^{n}}

Wende Exponentenregel an:

\frac{a ^{n}}{b ^{n}} = (\frac{a}{b})^{n}

\frac{2 ^{n}}{4 ^{n}} = (\frac{2}{4})^{n}

= n = 1 \sum \infty (\frac{2}{4})^{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{2}{4})^{n} - (\frac{2}{4})^{0}

n = 0 \sum \infty (\frac{2}{4})^{n} = 2

= 2 - (\frac{2}{4})^{0}

Vereinfache

= 1

x \to 0 lim (\frac{1 - cos ( 3 x )}{1 - cos ( 7 x )})

x \to 3 + lim (x^{2} - 9)

(x, y) \to (0, 2) lim (x^{3} y)

\frac{d}{d x} (40 x)

x \to 2 - lim (e^{\frac{3}{2 - x}})