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So I'm readnig a book on fundamental mathematics, it's not a calculus or analysis book so it barely introduces the concept of limit of sequences and then asks you to show some sequences and series converge. Because of this I don't really have much knowledge of this sort of thing and I would like some hints on how to solve this. Ideally the problem should be solved without using any advanced techniques, just the definitions of limit of series and sequences, since that's all the author expects you to have at that point.

First the problem asks to show that for any $x \in \mathbb{R}$ such that $0 < x < 1$, the sequence $x^0, x^1, x^2, ...$ converges to $0$. To do this, I simply showed that the series is decreasing and has a lower bound ($0$ for example) and thus it should converge to the maximum lower bound (we showed decreasing sequences behave like this on a previous exercise). Then I showed it cannot have a positive lower bound, so it must converge to $0$.

Now for part 2 the author asks to consider the series $x^0+x^1+x^2+...$ and show that for any $x \in \mathbb{R}$ such that $0 < x < 1$, the seires converges.

To my understanding what needs to be shown is that the sequence of partial sums converges, so $x^0, x^0+x^1, x^0+x^1+x^2,...$ converges. This sequence is clearly increasing, so if I can show that it is bounded above that should show it converges. My hunch is that it connverges to a number like $\frac{1}{1-x}$. But I can't justify that. Any hints?

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  • $\begingroup$ Can you simplify $$(x^0+x^1+\ldots+x^n)(x^0-x^1)?$$ Doing so should let you use the result of part 1. $\endgroup$ Jan 5, 2023 at 2:58
  • $\begingroup$ Critical that for any $x$ such that $0 < x < 1,$ you have that $$\lim_{n\to\infty} x^n = 0.$$ $\endgroup$ Jan 5, 2023 at 3:24
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    $\begingroup$ @BrianMoehring Thanks! That was the hints I needed :) So my hunch was right! it does converge to 1/1-x. I just came up with that because it worked on some specific cases like 1/2 and so on. $\endgroup$
    – zlaaemi
    Jan 5, 2023 at 3:33
  • $\begingroup$ @zlaaemi now that you've guessed what the series converges to, you can prove it rigorously (which it seems, based on the problem, you have to do) $\endgroup$
    – Alan Chung
    Jan 6, 2023 at 16:50

2 Answers 2

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One way you can reason about what the sequence converges to is the following (though from a mathematical point of view, the following argument isn't formal/rigorous).

Consider $$S = 1 + x + x^2 + x^3 + \dots,$$ $$Sx = x + x^2 + x^3 + \dots$$ Then what happens when you subtract the two equations and solve for $S$ (hint: a lot of terms should cancel)?

As a comment, the following example shows why this logic doesn't work if the series doesn't converge. Consider the infinite sum $$1-1+1-1+1-1+\dots$$ You can see that this series doesn't converge, since the definition of convergence is that the partial sums get infinitely close to some real number, but the partial sums of this series infinitely oscillates between 0 and 1, and thus doesn't converge. If you use the same argument as above, you get $$S = 1-1+1-1+1-1+1\dots$$ $$-S = -1+1-1+1-1+1\dots,$$ and if you subtract the two equations, it seems like all of the terms cancel and you get $$2S = 1 \Rightarrow S = \frac{1}{2},$$ which doesn't make sense.

So why did the method fail? Well, it's because the series doesn't converge (thus, the infinite sum is meaningless) and "subtracting" the two equations doesn't make sense because those equations are not valid in the first place! You can think of it like dividing both equations by 0. For example, suppose you have the equation $$x = 2x \Rightarrow x = 0.$$ But if you naively divided both sides by 0, then you get $$1 = 2,$$ which doesn't make sense because you did an illegal operation, and thus everything following the illegal operation is meaningless.

So in general, the method I proposed in this thread only works if you know a priori that the series converges, so that you are doing valid operations on an equation that has meaning.

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This can readily be shown by induction. For $n=0$ we have $\sum_{i=0}^n x^i = x^0=\frac{1-x^{0+1}}{1-x}$. Assume now that $\sum_{i=0}^n x^i = \frac{1-x^{n+1}}{1-x}$ for some nonnegative integer $n$, then $$ \sum_{i=0}^{n+1} x^i = x^{n+1}+\frac{1-x^{n+1}}{1-x} = \frac{1-x^{n+2}}{1-x}. $$ Since $\lim_{n\to\infty}x^n=0$, it follows that $$\lim_{n\to\infty}\sum_{i=0}^n x^i = \frac1{1-x}.$$

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