Please help me find the sum of this series:
1+23⋅12+23⋅56⋅122+23⋅56⋅89⋅123+⋯
All I could figure out was to find the nth term as:
an=2⋅(2+3)⋯(2+3(n−1))3⋅6⋅9⋯3(n−1)⋅12n−1
What To do ahead of it. I don't know. Please help.
Answer
Let S denote the sum. We write each term (with indices starting at 0) as
(n∏k=13k−13k)12n=∏n−1k=0(−23−k)n!(−12)n=(−2/3n)(−12)n.
Then we easily recognize S as a bionmial series and hence
S=∞∑n=0(−2/3n)(−12)n=(1−12)−2/3=22/3.
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