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

Lösung

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

\frac{1}{6}

Dezimale

0.16666 \dots

Schritte zur Lösung

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

n = 0 \sum \infty \frac{1}{3 ^{n}} = \frac{3}{2}

= \frac{3}{2} - \frac{1}{3 ^{0}} - \frac{1}{3 ^{1}}

Vereinfache

\frac{3}{2} - \frac{1}{3 ^{0}} - \frac{1}{3 ^{1}} : \frac{1}{6}

= \frac{1}{6}

x \to 0 - lim ((x + 5) e^{\frac{1}{x}})

x \to - 2 lim (x^{2})

\frac{d}{d x} (π^{e} + x^{e})

d er i v a t i v e e^{- \frac{1}{x}}

\int 7 x^{5} - 5 x^{4} + 6 x^{2} - 14 x + 3 d x