So...
$$S := \sum_{n=0}^{\infty} x^n=\frac{1}{1-x}\text{ for any } x<1$$
I just found out, and it tested it myself, that if this is true, then:
$$S_1 := \sum_{n=1}^{\infty} n = -\frac{1}{12}$$
Why? Because if the $S$ equality is true, then we derive $S$, and use $x=-1$, then $S_2 := 1-2+3-4+5-... = \frac{1}{4}$, and then $S_1-S_2 = 4\cdot S_1$, and $S_1 = -\frac{1}{12}$ which doesn't make any sense to me. I'm trying to see that everything is correct in this reasoning.
Though I could test $S$, because it is so for $x=\frac{1}{2}$, for $x=0$, for $x=-\frac{1}{2}$ (apparently), I can't test it for $-1$ or smaller numbers. So how do I know that $S = \sum_{n=1}^{\infty}x^n=\frac{1}{1-x}$ for any $x<1$? For my whole adult life I've known $S_1$ to be a divergent sequence. Now it converges? How?
Answer
It is true whenever $|x|<1,$ not whenever $x<1$.
As for how to show it, note that for any $x$ and any positive integer $n,$ we have $$\begin{align}(1-x)\sum_{k=0}^n x^k &= \sum_{k=0}^n(x^k-x^{k+1})\\ &= \left(\sum_{k=0}^n x^k\right)-\left(\sum_{k=0}^n x^{k+1}\right)\\ &= 1+\left(\sum_{k=1}^n x^k\right)-\left(x^{n+1}+\sum_{k=0}^{n-1}x^{k+1}\right)\\ &= 1+\left(\sum_{k=1}^n x^k\right)-\left(x^{n+1}+\sum_{k=1}^n x^k\right)\\ &= 1-x^{n+1}+\left(\sum_{k=1}^n x^k\right)-\left(\sum_{k=1}^n x^k\right)\\ &= 1-x^{n+1}.\end{align}$$ It is clear that $\sum\limits_{k=0}^\infty1^k$ diverges. Assuming that $x\neq 1,$ we may divide by the non-zero quantity $1-x$ to see that $$\sum_{k=0}^n x^k=\frac{1-x^{n+1}}{1-x}\tag{$\star$}$$ for all $x\ne1$ and all positive integers $n.$ In the case that $|x|\ge1,$ we have that $1-x^{n+1}$ diverges as $n$ grows without bound. Hence, the geometric series $\sum\limits_{k=0}^\infty x^k$ diverges by $(\star)$ and the $x=1$ case whenever $|x|\ge1.$ In the case that $|x|<1,$ we have that $1-x^{n+1}$ converges to $1$ as $n$ grows without bound. So, by $(\star),$ we have $$\sum_{k=0}^\infty x^k=\frac1{1-x}$$ whenever $|x|<1.$
No comments:
Post a Comment