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

Lösung

n = 1 \sum \infty 4 (\frac{π}{4})^{n}

4 (\frac{4}{4 - π} - 1)

Dezimale

14.63916 \dots

Schritte zur Lösung

n = 1 \sum \infty 4 (\frac{π}{4})^{n}

Wende die Regel der konstanten Multiplikation:

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

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

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

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

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

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

Vereinfache

= 4 (\frac{4}{4 - π} - 1)

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

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

n = 1 \sum \infty \frac{1}{3 n ^{4} + 12 n}

n = 1 \sum \infty \frac{4 n}{n ^{2} + 5}

n = 0 \sum \infty (\frac{5}{6})^{n - 1}