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I'm trying to learn how to build the generic series definition for a series of numbers. For some reason I'm having a hard time pulling out this pattern. I'm always pulling out the wrong details for when it comes to building the generic series definition. Is there a more controlled and efficient approach to determining these patterns.

What does it mean when a geometric series can be infinite or finite? It seems to me that you can only find the sum of a series that are finite or when r < |1|. However, I'm understanding that I can use the Maclaurin Series expansion formula to solve for f(x) when x is within range.

I'm just not understanding the application or idea of finding the sum vs. solving for f(x). Are these the same approaches? Am I doing the same thing but deriving a formula to calculate sums for infinite series?

Can someone elaborate on this as I'm having quite the time understanding how to approach these applications.

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Are you asking about how to find the general term of a series from a sum of particular numbers? \e.g., given $1 + \frac 12 + \frac 14 + \cdots$, knowing that the nth summand is given by $\frac 1{n^2}$, (simple example, I know, but is this what you mean by "pattern"?), or are you looking for a formula for $S(n)$: the $nth$ partial sum? –  amWhy May 22 '13 at 1:46
    
Is generic the same as geometric? You shift from one to the other. It sounds like geometric is correct. –  Ross Millikan May 22 '13 at 1:50
    
Correct, i'm asking about how to find the general term of a series like you mentioned. As to your question Ross I read that these series are called geometric series. As for "generic" I meant trying to find the general term of a series. This is all new material we are covering in my Calc class and I'm having a hard time understanding it so apologies if I come across unclear. –  Shane Yost May 22 '13 at 2:09
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1 Answer 1

A finite geometric series can always be summed. If the first term is $a$ and the ratio is $r$, so the series is $a, ar, ar^2, ar^3 \ldots ar^n$, the sum is $\frac {a(1-r^{n+1})}{1-r}$ [unless $r=1$, then it is $a(n+1)]$. When we let $n \to \infty$ this only converges if $|r| \lt 1$

For a MacLaurin series, the radius of convergence depends on how quickly the terms decrease. The above is the series for $\frac a{1-r}$. The series for $e^z$ converges for any $z$ in the complex plane because the terms decrease so quickly.

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