sum from n=0 to infinity of (2n-1)/(2^n)

Lösung

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

konvergiert

Schritte zur Lösung

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

Multipliziere aus

\frac{2 n - 1}{2 ^{n}} : 2^{1 - n} n - \frac{1}{2 ^{n}}

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

Wende die Summenregel an:

\sum a_{n} + b_{n} = \sum a_{n} + \sum b_{n}

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

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

konvergiert

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

= konvergiert - 2

= konvergiert

n = 1 \sum \infty \frac{10}{n ^{2}}

n = 1 \sum \infty \frac{1}{ln ( n + 2 )}

n = 1 \sum \infty \frac{n}{( n + 1 ) ^{2}}

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

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