I think you can see that .
So find the power series for and then multiply the entire series by .
Hi,
I'm asked to represent as a power series. Clearly it is similar to the geometric form .
I proceeded as follows: .
Of course this isn't in the form of a power series, but I don't really know where to go from here. The answer according to my text is - But I really don't know how they arrived at this - I tried to go backwards from the answer but ended up with something different from the original problem...so I don't know.
Thanks for the help ~ Might have a lot more questions on this topic hehe
I guess it's a misunderstanding - but I thought because the sigma operator behaves linearly you would distribute it across a quantity just like any other number, variable, etc. And in any case, I still don't see how they arrive at (by the way, with n = 1 not 0).
Ah! Thanks! I always felt like expanding out series was sort of a waste of time, but I'm starting to see more and more how necessary it is for identifying relationships & simplifications. For example, today I was reading about how Euler's formula is derived and thought it was very interesting how the power series expansion for evaluated at an exponential complex number allows you to immediately see the relationship between and with the exponential function after some quick factoring & separation.
Thanks again ~
Expanding helps sometimes but you have to be extremely careful. You have to make sure that the series you are expanding are originally convergent. If not, you may end up having wrong results that does not meet the definitions and theorems about series. For example, we all know that the series is divergent. But if you expand it and manipulate it you may end up having a convergent series whose value is 1/2!!
http://en.wikipedia.org/wiki/1-2%2B3-4
Good point. I suspect this same argument holds for some series that do converge as well! Conditionally convergent series that do not converge absolutely come to mind - for example, I had to prove on my most recent exam that given a conditionally convergent series, one can re-arrange its expansion in such a way to get any arbitrary value for big n. I succeeded =)