$ \sum_{n=1}^\infty x^n $For what values does $x$ converge? Diverge? Justify.

Question

For what values of $x$ does this converge? Diverge? And show justification $$ \sum_{n=1}^\infty x^n $$ I know that it converges when $|x| < 1$ and diverges when $|x| \ge 1$

However I don't know how to justify this. I can explain it by writing and I did it so by saying this

If you raise a number greater or equal to 1 to a power and add each value, it would go to infinity. However if you take a fraction and raise it to a power, the denominator gets larger and faster, than the numerator, meaning the last term will be $0$ and there will be a final term, meaning there will be a number of finite terms, rather than a number of infinite terms meaning it converges.

I think this is correct, however I'm not so sure how to justify this with "math"

The divergence for $|x| \geq 1$ can also be deduced from the $n$-th term test because the series converges only if $\lim\limits_{n \rightarrow \infty} x^n = 0$. (But this is not enough to show convergence when $|x| \lt 1$). — Tob Ernack, Jul 18 '17 at 04:17

score 2 · Accepted Answer · answered Jul 18 '17 at 02:59

2

HINT:

Recall that for $x\ne 1$

$$\sum_{n=1}^N x^n=\frac{x-x^{N+1}}{1-x}$$

answered Jul 18 '17 at 02:59

Mark Viola

184,670

I just got to that point right now! I substracted the two equations I created (the original one, and the one where i multiplied r into it) but now I'm unsure what to do from there – WildWombat Jul 18 '17 at 03:04
1

$S_N=1+x+\cdots x^N$, $xS_N=S_N+(x^{N+1}-1)$. Can you finish now? – Mark Viola Jul 18 '17 at 03:07
I did it, but this is where I ended up – WildWombat Jul 18 '17 at 03:31
You have $S_N=\frac{x-x^{N+1}}{1-x}$. As $N\to \infty$, $S_N\to \frac{x}{1-x}$ for $|x|<1$ and diverges for $|x|\ge 1$, which is evident by examining $\lim_{N\to \infty}x^{N+1}$. – Mark Viola Jul 18 '17 at 03:40
This makes so much sense now, thank you so much! – WildWombat Jul 18 '17 at 04:33
Pleased to hear. And you're welcome. My pleasure. – Mark Viola Jul 18 '17 at 13:31

score 1 · Answer 2 · answered Jul 18 '17 at 03:06

1

One can also use the ratio test to show absolute convergence: $$ \lim_{n \to \infty} \left| \frac{a^{n+1}}{a^n} \right| < 1 $$

$$ \lim_{n \to \infty} \left| \frac{x^{n+1}}{x^n} \right| = \lim_{n\to\infty}|x| = |x| < 1 $$

answered Jul 18 '17 at 03:06

Dando18

5,472

I'm a bit worried this might be circular. Proofs of the ratio test usually involve bounding $\sum |a_n|$ by a convergent geometric series. – MathematicsStudent1122 Jul 18 '17 at 03:13
@MathematicsStudent1122 I believe the answers here avoid referencing convergent geometrics using an $\epsilon$ proof. – Dando18 Jul 18 '17 at 03:21

$ \sum_{n=1}^\infty x^n $For what values does $x$ converge? Diverge? Justify.

2 Answers2