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Sum of Infinite Series | Calculate the Sum of a Geometric Series
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[QUOTE="RPinPA, post: 6089041, member: 651116"] No, it's not a geometric series. The characteristic of a geometric series is that the ratio between successive terms is a constant. The ratio ##a_{n+1}/a_n## in this case is ##(n 4^n) / [(n+1)4^{n+1}]## or ##n/[4(n+1)]##, definitely not a constant. That said, I'm not sure off-hand what trick might work here. Here's a PDF that shows some of the tricks for transforming infinite series into something that you can work with. [URL='http://web.math.ucsb.edu/~cmart07/Evaluating%20Series.pdf']http://web.math.ucsb.edu/~cmart07/Evaluating Series.pdf[/URL] I think the basic hope here would be if you can use the Taylor series approach: Can you transform this into some known Taylor series, evaluated at a particular value? Then it would be that function at that value. Or perhaps the derivative of a known Taylor series? That's where I'd start looking, but I have no specific advice to give you. Edit: I think I found one. Look at a table of common Taylor series and find one where the [I]n[/I]-th term has an [I]n[/I] in the denominator. You can make that work for you. That is, the [I]n[/I]-th term of the expansion of ##f(x)## can be the same as the [I]n[/I]-th term of this series, for a particular [I]x[/I]. [/QUOTE]
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Sum of Infinite Series | Calculate the Sum of a Geometric Series
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