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I'm looking for help on how to find the sum and interval of convergence of this series (Starts at 0 and goes to infinity). Now this one is giving me trouble because I've never seen a series with the "α" before. I was thinking maybe I could start by comparing with $\sum x^n$ which has the sum of $\frac{1}{1-x}$, but I am not sure.

$$\sum_{n=0}^{+\infty}(\alpha x)^n, \alpha>0$$

Thanks for your help!

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  • $\begingroup$ How about $1/(1-\alpha x)$ ? $\endgroup$ – Pierpaolo Vivo Apr 16 '16 at 21:22
  • $\begingroup$ @PierpaoloVivo thats only true for $ |\alpha x|<1$ $\endgroup$ – daniels_pa Apr 16 '16 at 21:26
  • $\begingroup$ Is there more information we know about the numbers in this question? $\endgroup$ – Prince M Apr 16 '16 at 21:30
  • $\begingroup$ Yeah, as in @daniels_pa 's answer, note that it is a geometric series, so you can use the usual formulas for $S_∞$ $\endgroup$ – KR136 Apr 16 '16 at 21:44
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Yes the sum you wrote

$$ \sum_{n=0}^{+\infty} x^n $$

Converges for $|x|<1$

Same can be applied here.
For $|\alpha x|<1$ we have that

$$ \sum_{n=0}^{+\infty} (\alpha x)^n = \frac{1}{1-\alpha x} $$

So it converges when $-1<\alpha x<1$ . There we have

For
$\alpha=0$, converges $\forall x\in \mathbb{R}$
$\alpha>0$, converges $\forall x\in (-\frac{1}{\alpha},\frac{1}{\alpha})$
$\alpha<0$, converges $\forall x\in (\frac{1}{\alpha},-\frac{1}{\alpha})$

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  • $\begingroup$ Thanks for the reply! I have to be honest though, I don't know what the upside-down A or the 'E' symbols mean. My math history has been: College Algebra > Trig > Pre-cal > Cal1 > Cal2. $\endgroup$ – Python guy Apr 16 '16 at 22:51
  • $\begingroup$ Upside-down A means for all and the 'E' symbol or '$\in$' means belongs to as you can see here $\endgroup$ – daniels_pa Apr 16 '16 at 23:03
  • $\begingroup$ Just an added note that may help you understand, the reason why this works is a simple substitution with $y = \alpha x$ and then the sum @daniels_pa wrote goes to $\frac{1}{1-y} = \frac{1}{1-\alpha x} $. You may want to write this in your formal proof as an intermediate step just for clarity. $\endgroup$ – user244643 Apr 17 '16 at 4:22

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