(PDE) (Uniform Series Convergence) I'm trying to figure out formally how the nth partial sum of the following series was determined; I can intuitively see it described as a couple of answers, but I can't figure out why this particular one is the "right" one.

Question

mendicant_bias · Answer

I can see that if someone taking some partial sum of this series chose to "stop" at +x^n as opposed to stopping at -x^n, your nth partial sum would just be one. Why isn't this equally valid?

mendicant_bias · Answer

@SithsAndGiggles

mendicant_bias · Answer

I don't think I remember how to formally determine the nth partial sum anyways, but-yeah-nth partial sum is $$S_n(x)=1-x+x-x^2+x^2+-...-x^{n-1}+x^{n-1}-x^n=1-x^n$$

mendicant_bias · Answer

What I don't understand is, say I try to "write out" each term of the nth partial sum in a similar way, and say I choose to stop on the third "type" of term, where it's +x^n; that +x^n would cancel out with the -x^n term, and you'd be left with the nth partial sum equalling one. Why is this formally wrong? I get that there's no use coming to the conclusion that a series like this behaves in that way, it totally misses the point, but it isn't "wrong" from what I know. e.g. does the series have two possible nth partial sums, and if not, why is this one right?

mendicant_bias · Answer

It also confuses me, I guess, that if you use the formal definition of the nth partial sum and have to produce every term from the original sequence as needed with n, wouldn't you get 1, then? I'll introduce the original general sequence term for clarification: 

$$\sum_{n=1}^{\infty}x^{n-1}(1-x)$$

anonymous · Answer

Judging by the partial sum you've provided, it looks like the sum is
$$S_n(x)=\sum_{k=0}^n (x^k-x^{k+1})$$
so you have
$$S_1(x)=\sum_{k=0}^1 (x^k-x^{k+1})=(1-x)+(x-x^2)=1-x^2$$
and so on up to
$$S_n(x)=\sum_{k=0}^n (x^k-x^{k+1})=1-x^n$$

mendicant_bias · Answer

Okay, yeah, nevermind, that makes sense, I didn't look much at all at the original series and, looking from the nth partial sum backwards, I imagined a different series term that made no sense in the way I was thinking about it.

anonymous · Answer

Right, the given series telescopes to the final expression.\]

blurbendy · Answer

"Mendicant_Bias" is my favorite username =)

mendicant_bias · Answer

Lol, thanks, heh

mendicant_bias · Answer

Oh yeah, I had another question regarding this series which I don't understand, which is that my book states$$S(x)=\lim_{n\rightarrow\infty}S_n(x)=1$$

mendicant_bias · Answer

Oh, nevermind, had to do with the bounds/region involved