Answer to Question #189971 in Calculus for Moel Tariburu

Question #189971

Find the sum of the infinite series ∑_(n=1)^(+∞)▒2^(k+1)/3^(k+2) 


1
Expert's answer
2021-05-11T14:45:40-0400

First let us rewrite it as

k12k+13k+2=232k12k3k=29k1(23)k\sum_{k\geq 1} \frac{2^{k+1}}{3^{k+2}}=\frac{2}{3^2}\sum_{k\geq 1} \frac{2^k}{3^k} = \frac{2}{9} \sum_{k\geq 1} (\frac{2}{3})^k

The last series is a geometric series, for which we have

k1qk=q1q\sum_{k\geq 1} q^k = \frac{q}{1-q} (which we can deduce from k=1nqk=qqn+11q\sum_{k=1}^n q^k = \frac{q-q^{n+1}}{1-q} and qn+10q^{n+1}\to 0 for q<1|q|<1) and therefore the result is given by

29k1(23)k=29×23123=49\frac{2}{9} \sum_{k\geq 1} (\frac{2}{3})^k =\frac{2}{9} \times \frac{\frac{2}{3}}{1-\frac{2}{3}}=\frac{4}{9}


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