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HallsofIvy
#6
Jun12-05, 10:02 AM
Math
Emeritus
Sci Advisor
Thanks
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Quote Quote by steven187
so your sayin that it is possible in some cases to derive a formula for a series besides a geometric series but in most cases its not possible? , and about that series it was suppose to be

[tex]\sum_{n=1}^{\infty}\frac {(-2)^{-n}}{n+1}[/tex]
like would i be able to derive a formula for this series? or how would i be able to find the sum of such a series?
Some times you can recognize a series as a special case of a Taylor's series for a function. To sum the particular series you give here, I would recall that the Taylor's series for ln(x+1) is [tex]\sum_{n=1}^{\infty}(-1)^n \frac{x^n}{n}[/tex] (and converges for -1< x< 1). That differs from your series on having n+1 in the denominator: Okay, change the numbering slightly. If we let i= n+1 then n= i-1 and your series becomes [tex]\sum_{i=2}^{\infty}\frac{(-2)^{-i+1}}{i}[/tex]. I can just take that (-2)1 out of the entire series: [tex](-2)\sum_{i=2}^{\infty}\frac{(-2)^{-i}}{i}[/tex]. We're almost there! Since I need (-2)-i instead of xi, take x= -1/2 which is in the radius of convergence. The fact that the sum now starts at i= 2 instead of 1 is not a problem: just calculate that separately- when i= 0 the term is (-2)0/1= -2: Your sum is [tex](-2)\sum{i=1}\frac{(-1)^i}{i}+ 2)= (-2)ln(-1/2+ 1)+ 2= -2ln(1/2)+ 2= 2+ 2 ln(2)= 2(1+ ln(2))[/tex].